An historic problem in the history of math was to find the area under the graph of $f(x)=x^2$ between $[0,1]$.

There isn't a ready-made formula for the area of this shape, as there is for a triangle or a square. However, Archimedes found a method to compute areas enclosed by a parabola and line segments that cross the parabola.

The figure illustrates a means to compute the area bounded by the parabola, the line $y=1$ and the line $x=0$ using triangles. It suggests that this area can be found by adding the following sum

\[ ~ A = 1/2 + 1/8 + 2 \cdot (1/8)^2 + 4 \cdot (1/8)^3 + \cdots ~ \]

This value is $2/3$, so the area under the curve would be $1/3$. Forget about this specific value - which through more modern machinery becomes uneventful - and focus for a minute on the method: a problem is solved by a suggestion of an infinite process, in this case the creation of more triangles to approximate the unaccounted for area. This is the so-call method of exhaustion known since the 5th century BC. Archimedes used this method to solve a wide range of area problems related to basic geometric shapes, including a more general statement of what we described above.

The $\cdots$ in the sum expression are the indication that this process continues and that the answer is at the end of an *infinite* process. To make this line of reasoning rigorous requires the concept of a limit. The concept of a limit is then an old one, but it wasn't until the age of calculus that it was formalized.

Fermat in the 1600s essentially took a limit to find the slope of a tangent line to a polynomial curve. Newton in the late 1600s, exploited the idea in his development of calculus (as did Leibniz). Yet it wasn't until the 1800s that Bolzano, Cauchy and Weierstrass put the idea on a firm footing.

To make things more precise, we begin by discussing the limit of a univariate function as $x$ approaches $c$.

Informally, if a limit exists it is the value that $f(x)$ gets close to as $x$ gets close to - but not equal to - $c$.

The modern formulation is due to Weirstrass:

The limit of $f(x)$ as $x$ approaches $c$ is $L$ if for every real $\epsilon > 0$, there exists a real $\delta > 0$ such that for all real $x$, $0 < \lvert x âˆ’ c \rvert < \delta$ implies $\lvert f(x) âˆ’ L \rvert < \epsilon$. The notation used is $\lim_{x \rightarrow c}f(x) = L$.

We comment on this later.

Cauchy begins his incredibly influential treatise on calculus considering two examples, the limit as $x$ goes to $0$ of

\[ ~ \frac{\sin(x)}{x} \quad\text{and}\quad (1 + x)^{1/x}. ~ \]

These take the indeterminate forms $0/0$ and $1^\infty$, which are found by just putting $0$ in for $x$. An expression does not need to be defined at $c$, as these two aren't, to discuss its limit. Cauchy illustrates two methods to approach the questions above. The first is to pull out an inequality:

\[ ~ \frac{\sin(x)}{\sin(x)} > \frac{\sin(x)}{x} > \frac{\sin(x)}{\tan(x)}. ~ \]

This bounds the expression $\sin(x)/x$ between $1$ and $\cos(x)$ and as $x$ gets close to $0$, the value of $\cos(x)$ "clearly" goes to $1$, hence $L$ must be $1$. This is an application of the squeeze theorem.

To discuss the case of $(1+x)^{1/x}$, Cauchy first resorts to log tables to illustrate an approximate value. We can use `Julia`

to find his approximate value:

x = 1/10000 (1 + x)^(1/x)

2.7181459268249255

A table can show the progression to this value:

f(x) = (1 + x)^(1/x) xs = [1/10^i for i in 1:5] [xs f.(xs)]

5Ã—2 Array{Float64,2}: 0.1 2.59374 0.01 2.70481 0.001 2.71692 0.0001 2.71815 1.0e-5 2.71827

However viewed, this is an approximate value for the actual answer - which is $e$. Cauchy demonstrates this value with a fair amount of work.

These two cases illustrate that though the definition of the limit exists, the computation of a limit is generally found by other means and the intuition of the value of the limit can be gained numerically.

First it should be noted that for most of the functions encountered, the concepts of a limit at a typical point $c$ is nothing more than just function evaluation at $c$. This is because, at a typical point, the functions are nicely behaved. However, most questions asked about limits find points that are not typical. For these, the result of evaluating the function at $c$ is typically undefined, and the value comes in one of several *indeterminate forms*: $0/0$, $\infty/\infty$, $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, and $\infty^0$.

`Julia`

can help - at times - identify these indeterminate forms, as many such operations produce `NaN`

. For example:

0/0, Inf/Inf, 0 * Inf, Inf - Inf

(NaN, NaN, NaN, NaN)

However, the values with powers generally do not help, as the IEEE standard has `0^0`

evaluating to 1:

0^0, 1^Inf, Inf^0

(1, 1.0, 1.0)

However, this can be unreliable, as floating point issues may mask the true evaluation. However, as a cheap trick it can work. So, the limit as $x$ goes to $1$ of $\sin(x)/x$ is simply found by evaluation:

x = 1 sin(x) / x

0.8414709848078965

But at $x=0$ we get an indicator that there is an issue with just evaluating the function:

x = 0 sin(x) / x

NaN

The above is really just a heuristic. For some functions this is just not true. For example, the $f(x) = \sqrt{x}$ is only defined on $[0, \infty)$ There is technically no limit at $0$, per se, as the function is not defined around $0$. Other functions jump at values, and will not have a limit, despite having well defined values. The `floor`

function is the function that rounds down to the nearest integer. At integer values there will be a jump, even though the function is defined.

Let's return to the function $f(x) = \sin(x)/x$. This function was studied by Euler as part of his solution to the Basel problem. He knew that near $0$, $\sin(x) \approx x$, so the ratio is close to $1$ if $x$ is near $0$. Hence, the intuition is $\lim_{x \rightarrow 0} \sin(x)/x = 1$, as Cauchy wrote. We can verify this limit graphically two ways. First, a simple graph shows no issue at $0$:

using CalculusWithJulia # loads SymPy using Plots f(x) = sin(x)/x plot(f, -pi/2, pi/2)

The $y$ values of the graph seem to go to $1$ as the $x$ values get close to 0. (That the graph looks defined at $0$ is due to the fact that the points sampled to graph do not include $0$. Otherwise, since `f(0)`

is `NaN`

, there would be no lines connecting to the value with `x=0`

.)

We can also verify Euler's intuition through this graph:

plot([sin, x -> x], -pi/2, pi/2)

That the two are indistinguishable near $0$ makes it easy to see that their ratio should be going towards $1$.

The graphical approach to limits - plotting $f(x)$ around $c$ and observing if the $y$ values seem to converge to an $L$ value when $x$ get close to $c$ - allows us to gather quickly if a function seems to have a limit at $c$, though the precise value of $L$ may be hard to identify.

Consider now the following limit

\[ ~ \lim_{x \rightarrow 2} \frac{x^2 - 5x + 6}{x^2 +x - 6} ~ \]

Noting that this is a ratio of nice polynomial functions, we first check whether there is anything to do:

f(x) = (x^2 - 5x + 6) / (x^2 + x - 6) c = 2 f(c)

NaN

The `NaN`

indicates that this function is indeterminate at $c=2$. A quick plot gives us an idea that the limit exists and is roughly $-0.2$:

c, delta = 2, 1 plot(f, c - delta, c + delta)

The graph looks "continuous." In fact, the value $c=2$ is termed a *removable singularity* as redefining $f(x)$ to be $-0.2$ when $x=2$ results in a "continuous" function.

As an aside, we can redefine `f`

using the "ternary operator":

f(x) = x == 2.0 ? -0.2 : (x^2 - 5x + 6) / (x^2 + x - 6)

f (generic function with 1 method)

This particular case is a textbook example: one can easily factor $f(x)$ to get:

\[ ~ f(x) = \frac{(x-2)(x-3)}{(x-2)(x+3)} ~ \]

Written in this form, we clearly see that this is the same function as $g(x) = (x-3)/(x+3)$ when $x \neq 2$. The function $g(x)$ is "continuous" at $x=2$. So were one to redefine $f(x)$ at $x=2$ to be $g(2) = (2 - 3)/(2 + 3) = -0.2$ it would be made continuous, hence the term removable singularity.

The investigation of $\lim_{x \rightarrow 0}(1 + x)^{1/x}$ by evaluating the function at $1/10000$ by Cauchy can be done much more easily nowadays. As does a graphical approach, a numerical approach can give insight into a limit and often a good numeric estimate.

The basic idea is to create a sequence of $x$ values going towards $c$ and then investigate if the corresponding $y$ values are eventually near some $L$.

Best, to see by example. Suppose we are asked to investigate

\[ ~ \lim_{x \rightarrow 25} \frac{\sqrt{x} - 5}{\sqrt{x - 16} - 3}. ~ \]

We first define a function and check if there are issues at 25:

f(x) = (sqrt(x) - 5) / (sqrt(x-16) - 3) c = 25 f(c)

NaN

So yes, an issue of the indeterminate form $0/0$. We investigate numerically by making a set of numbers getting close to $c$. This is most easily done making numbers getting close to $0$ and adding them to or subtracting them from $c$. Some natural candidates are negative powers of 10:

hs = [1/10^i for i in 1:8]

8-element Array{Float64,1}: 0.1 0.01 0.001 0.0001 1.0e-5 1.0e-6 1.0e-7 1.0e-8

We can add these to $c$ and then evaluate:

xs = c .+ hs ys = f.(xs)

8-element Array{Float64,1}: 0.6010616008415922 0.6001066157341047 0.6000106661569936 0.6000010666430725 0.6000001065281493 0.6000000122568625 0.5999999946709295 0.6

To visualize, we can put in a table using `[xs ys]`

notation:

[xs ys]

8Ã—2 Array{Float64,2}: 25.1 0.601062 25.01 0.600107 25.001 0.600011 25.0001 0.600001 25.0 0.6 25.0 0.6 25.0 0.6 25.0 0.6

The $y$-values seem to be getting near $0.6$.

Since limits are defined by the expression $0 < \lvert x-c\rvert < \delta$, we should also look at values smaller than $c$. There isn't much difference (note the `.-`

sign in `c .- hs`

):

xs = c .- hs ys = f.(xs) [xs ys]

8Ã—2 Array{Float64,2}: 24.9 0.598928 24.99 0.599893 24.999 0.599989 24.9999 0.599999 25.0 0.6 25.0 0.6 25.0 0.6 25.0 0.6

Same story. The numeric evidence supports a limit of $L=0.6$.

Let $f(x) = x^x$ and consider the ratio:

\[ ~ \frac{f(c + h) - f(c)}{h} ~ \]

As $h$ goes to $0$, this will take the form $0/0$ in most cases, and in the particular case of $f(x) = x^x$ and $c=1$ it will be. The expression has a geometric interpretation of being the slope of the secant line connecting the two points $(c,f(c))$ and $(c+h, f(c+h))$.

To look at the limit in this example, we have (recycling the values in `hs`

):

c = 1 f(x) = x^x ys = [(f(c + h) - f(c)) / h for h in hs] [hs ys]

8Ã—2 Array{Float64,2}: 0.1 1.10534 0.01 1.01005 0.001 1.001 0.0001 1.0001 1.0e-5 1.00001 1.0e-6 1.0 1.0e-7 1.0 1.0e-8 1.0

The limit looks like $L=1$. A similar check on the left will confirm this numerically:

ys = [(f(c + h) - f(c)) / h for h in -hs] [-hs ys]

8Ã—2 Array{Float64,2}: -0.1 0.904674 -0.01 0.99005 -0.001 0.999 -0.0001 0.9999 -1.0e-5 0.99999 -1.0e-6 0.999999 -1.0e-7 1.0 -1.0e-8 1.0

The numeric approach often gives a good intuition as to the existence of a limit and its value. However, it can be misleading. Consider this limit question:

\[ ~ \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^2}. ~ \]

We can see that it is indeterminate of the form $0/0$:

f(x) = (1 - cos(x)) / x^2 f(0)

NaN

What is the value of $L$, if it exists? A quick attempt numerically yields:

c = 0 xs = c .+ hs ys = [f(x) for x in xs] [xs ys]

8Ã—2 Array{Float64,2}: 0.1 0.499583 0.01 0.499996 0.001 0.5 0.0001 0.5 1.0e-5 0.5 1.0e-6 0.500044 1.0e-7 0.4996 1.0e-8 0.0

Hmm, the values in `ys`

appear to be going to $0.5$, but then end up at $0$. Is the limit $0$ or $1/2$? The answer is $1/2$. The last $0$ is an artifact of floating point arithmetic. To investigate, we look more carefully at the two ratios:

y1s = [1 - cos(x) for x in xs] y2s = [x^2 for x in xs] [xs y1s y2s]

8Ã—3 Array{Float64,2}: 0.1 0.00499583 0.01 0.01 4.99996e-5 0.0001 0.001 5.0e-7 1.0e-6 0.0001 5.0e-9 1.0e-8 1.0e-5 5.0e-11 1.0e-10 1.0e-6 5.00044e-13 1.0e-12 1.0e-7 4.996e-15 1.0e-14 1.0e-8 0.0 1.0e-16

Looking at the bottom of the second column reveals the error. The value of `1 - cos(1.0e-8)`

is `0`

and not a value around `5e-17`

, as would be expected from the pattern above it. This is because the smallest floating point value less than `1.0`

is more than `5e-17`

units away, so `cos(1e-8)`

is evaluated to be `1.0`

. There just isn't enough granularity to get this close to $0$.

Not that we needed to. The answer would have been clear if we had stopped with `x=1e-6`

, say.

In general, some functions will frustrate the numeric approach. It is best to be wary of results. At a minimum they should confirm what a quick graph shows.

The `SymPy`

package provides a `limit`

function for finding the limit of an expression in a given variable. It is loaded above, when the `CalculusWithJulia`

package was loaded. The `limit`

function's use requires the expression, the variable and a value for $c$. (Similar to the three things in the notation $\lim_{x \rightarrow c}f(x)$.)

For example, the limit at $0$ of $(1-\cos(x))/x^2$ is easily handled:

@vars x real=true f(x) = (1 - cos(x)) / x^2 limit(f(x), x=>0) # f(x) is a symbolic expression when x is

\begin{equation*}\frac{1}{2}\end{equation*}

The pair notation (`x=>0`

) is used to indicate the variable and the value it is going to.

As a convenience, there is also an "operator" version of `limit`

for univariate functions, where only a function object and `c`

need be specified:

limit(f, 0)

\begin{equation*}\frac{1}{2}\end{equation*}

Find the limits:

\[ ~ \lim_{x \rightarrow 0} \frac{2\sin(x) - \sin(2x)}{x - \sin(x)}, \quad \lim_{x \rightarrow 0} \frac{e^x - 1 - x}{x^2}, \quad \lim_{\rho \rightarrow 0} \frac{x^{1-\rho} - 1}{1 - \rho}. ~ \]

We have for the first:

limit( (2sin(x) - sin(2x)) / (x - sin(x)), x=>0)

\begin{equation*}6\end{equation*}

The second is similarly done, though here we define a function for variety:

f(x) = (exp(x) - 1 - x) / x^2 limit(f, 0)

\begin{equation*}\frac{1}{2}\end{equation*}

Finally, for the third we define a new variable and proceed:

@vars rho real=true limit( (x^(1-rho) - 1) / (1 - rho), rho=>1)

\begin{equation*}\log{\left(x \right)}\end{equation*}

The algorithm implemented in `SymPy`

for symbolic limits is quite powerful. However, some care must be exercised to avoid undesirable conversions from exact values to floating point values.

To Illustrate, let's look at the limit as $x$ goes to $\pi/2$ of $f(x) = \cos(x) / (x - \pi/2)$. We follow our past practice:

c = pi/2 f(x) = cos(x) / (x - pi/2) f(c)

Inf

The value is not `NaN`

, rather `Inf`

. This is because `cos(pi/2)`

is not exactly $0$ as it should be, as `pi/2`

is rounded. This minor difference is important. If we try and correct for this by using `PI`

we have:

limit(f(x), x=>PI/2)

\begin{equation*}0\end{equation*}

The value is not right, as this simple graph suggests the limit is in fact $-1$:

plot(f, c - pi/4, c + pi/4)

The difference between `pi`

and `PI`

can be significant, and though usually `pi`

is silently converted to `PI`

, it doesn't happen here as the division by `2`

happens first, which turns the symbol into an approximate floating point number. Hence, `SymPy`

is giving the correct answer for the problem it is given, it just isn't the problem we wanted to look at.

Trying again, being more aware of how `pi`

and `PI`

differ, we have:

f(x) = cos(x) / (x - PI/2) limit(f(x), x => PI/2)

\begin{equation*}-1\end{equation*}

The `limit`

function doesn't compute limits from the definition, rather it applies some known facts about functions within a set of rules. Some of these rules are the following. Suppose the individual limits of $f$ and $g$ always exist below.

Rule | Description |
---|---|

$\lim_{x \rightarrow c} (a \cdot f(x) + b \cdot g(x)) = a \cdot
\lim_{x \rightarrow c} f(x) + b \cdot \lim_{x \rightarrow c}
g(x)$
| This says that limits involving sums, differences or scalar multiples of functions can be computed by first doing the individual limits and then combining the answers. |

$\lim_{x \rightarrow c} f(x) \cdot g(x) = \lim_{x \rightarrow c}
f(x) \cdot \lim_{x \rightarrow c} g(x)$
| This says limits of products can be found by computing the limit of the individual factors and then combining. |

$\lim_{x \rightarrow c} f(x) / g(x) = \lim_{x \rightarrow c} f(x) / \lim_{x \rightarrow c} g(x)$ - provided $\lim_{x \rightarrow c} g(x) \neq 0$ | This says limits of ratios can be found by computing the limit of the individual terms and then dividing provided you don't divide by $0$. The last part is really important, as this rule is no help with the common indeterminate form $0/0$. |

$\lim_{x \rightarrow c} (f \circ g)(x) = \lim_{x \rightarrow L} f(x)$, where $\lim_{x \rightarrow c}g(x) = L$ | This says the limit of compositions can be found by taking the limit of the interior function ($L$) and then finding the limit of the exterior function as $x$ approaches $L$. |

These, together with the fact that our basic algebraic functions have limits that can be found by simple evaluation, mean that many limits are easy to compute.

For example, consider for some non-zero $k$ the following limit:

\[ ~ \lim_{x \rightarrow 0} \frac{\sin(kx)}{x}. ~ \]

This is clearly related to the function $f(x) = \sin(x)/x$, but is in fact the limit of the function $g(x) = k f(kx)$. This then follows:

\[ ~ \lim_{x \rightarrow 0} \frac{\sin(kx)}{x} = \lim_{x \rightarrow 0} k f(kx) = k \lim_{x \rightarrow 0} f(kx) = k\lim_{x \rightarrow 0} f(x) = k. ~ \]

Why does the limit of $f(kx)$ at 0 equal the limit of $f(x)$ at 0? Well, $f(kx)$ is a composition, $f(h(x))$ where $h(x) = kx$. The limit of $h$ at 0 is 0, so this is a special case of the rule for compositions.

Basically when taking a limit as $x$ goes to $0$ we can multiply $x$ by any constant and figure out the limit for that. (It is as though we "go to" $0$ faster or slower. but are still going to $0$. This can be generalized to any function $g(x)$ with a limit of $0$ at $0$: $\lim_{x \rightarrow 0}f(g(x)) = \lim_{x \rightarrow 0}f(x)$.)

Consider this complicated limit found on this Wikipedia page.

\[ ~ \lim_{x \rightarrow 1/2} \frac{\sin(\pi x)}{\pi x} \cdot \frac{\cos(\pi x)}{1 - (2x)^2}. ~ \]

We know the first factor has a limit found by evaluation: $2/\pi$, so it is really just a constant. The second we can compute:

g(x) = cos(PI*x) / (1 - (2x)^2) limit(g, 1//2)

\begin{equation*}0.25 \pi\end{equation*}

Putting together, we would get $1/2$. Which we could have done directly in this case:

limit(sin(PI*x)/(PI*x) * g(x), x=>1//2)

\begin{equation*}0.5\end{equation*}

Consider again the limit of $\cos(\pi x) / (1 - (2x)^2)$ at $c=1/2$. A graph of both the top and bottom functions shows the indeterminate, $0/0$, form:

plot(cos(pi*x), 0.4, 0.6) plot!(1 - (2x)^2)

However, following Euler's insight that $\sin(x)/x$ will have a limit at $0$ of $1$ as $\sin(x) \approx x$, and $x/x$ has a limit of $1$ at $c=0$, we can see that $\cos(\pi x)$ looks like $-\pi\cdot (x - 1/2)$ and $(1 - (2x)^2)$ looks like $-4(x-1/2)$ around $x=1/2$:

plot(cos(pi*x), 0.4, 0.6) plot!(-pi*(x - 1/2))

plot(1 - (2x)^2, 0.4, 0.6) plot!(-4(x - 1/2))

So around $c=1/2$ the ratio should look like $-\pi (x-1/2) / ( -4(x - 1/2)) = \pi/4$, which indeed it does, as that is the limit.

This is the basis of L'HoÌ‚pital's rule, which we will return to once the derivative is discussed.

The formal definition of a limit involves clarifying what it means for $f(x)$ to be "close to $L$" when $x$ is "close to $c$". These are quantified by the inequalities $0 < \lvert x-c\rvert < \delta$ and the $\lvert f(x) - L\rvert < \epsilon$. The second does not have the restriction that it is greater than $0$, as indeed $f(x)$ can equal $L$. The order is important: it says for any idea of close for $f(x)$ to $L$, an idea of close must be found for $x$ to $c$.

The key is identifying a value for $\delta$ for a given value of $\epsilon$.

A simple case is the linear case. Consider the function $f(x) = 3x + 2$. Verify that the limit at $c=1$ is $5$.

We show "numerically" that $\delta = \epsilon/3$.

f(x) = 3x + 2 c, L = 1, 5 epsilon = rand() # some number in (0,1) delta = epsilon / 3 xs = c .+ delta * rand(100) # 100 numbers, c < x < c + delta as = [abs(f(x) - L) < epsilon for x in xs] all(as) # are all the as true?

true

These lines produce a random $\epsilon$, the resulting $\delta$, and then verify for 100 numbers within $(c, c+\delta)$ that the inequality $\lvert f(x) - L \rvert < \epsilon$ holds for each. Running them again and again should always produce `true`

if $L$ is the limit and $\delta$ is chosen properly.

(Of course, we should also verify values to the left of $c$.)

If there is worry about the possibility that `epsilon == 0`

, we could replace its definition with `epsilon = max(eps(), rand())`

.

In this case, $\delta$ is easy to guess, as the function is linear and has slope $3$. This basically says the $y$ scale is 3 times the $x$ scale. For non-linear functions, finding $\delta$ for a given $\epsilon$ can be a challenge. For the function $f(x) = x^3$, illustrated below, a value of $\delta=\epsilon^{1/3}$ is used for $c=0$:

From the graph, find the limit:

\[ ~ L = \lim_{x\rightarrow 1} \frac{x^2âˆ’3x+2}{x^2âˆ’6x+5} ~ \]

From the graph, find the limit $L$:

\[ ~ L = \lim_{x \rightarrow -2} \frac{x}{x+1} \frac{x^2}{x^2 + 4} ~ \]

Graphically investigate the limit

\[ ~ L = \lim_{x \rightarrow 0} \frac{e^x - 1}{x}. ~ \]

What is the value of $L$?

Graphically investigate the limit

\[ ~ \lim_{x \rightarrow 0} \frac{\cos(x) - 1}{x}. ~ \]

The limit exists, what is the value?

The following limit is commonly used:

\[ ~ \lim_{h \rightarrow 0} \frac{e^{x + h} - e^x}{h} = L. ~ \]

Factoring out $e^x$ from the top and using rules of limits this becomes,

\[ ~ L = e^x \lim_{h \rightarrow 0} \frac{e^h - 1}{h}. ~ \]

What is $L$?

The following limit is commonly used:

\[ ~ \lim_{h \rightarrow 0} \frac{\sin(x + h) - \sin(x)}{h} = L. ~ \]

The answer should depend on $x$, though it is possible it is a constant. Using a double angle formula and the rules of limits, this can be written as:

\[ ~ L = \cos(x) \lim_{h \rightarrow 0}\frac{\sin(h)}{h} + \sin(x) \lim_{h \rightarrow 0}\frac{\cos(h)-1}{h}. ~ \]

Using the last result, what is the value of $L$?

Let's look at the function $f(x) = x \sin(1/x)$. A graph around $0$ can be made with:

f(x) = x == 0 ? NaN : x * sin(1/x) c, delta = 0, 1/4 plot([f, abs, x -> -abs(x)], c - delta, c + delta)

This graph clearly oscillates near $0$. To the graph of $f$, we added graphs of both $g(x) = \lvert x\rvert$ and $h(x) = - \lvert x\rvert$. From this graph it is easy to see by the "squeeze theorem" that the limit at $x=0$ is $0$. Why?

Find the limit as $x$ goes to $2$ of

\[ ~ f(x) = \frac{3x^2 - x -10}{x^2 - 4} ~ \]

Find the limit as $x$ goes to $-2$ of

\[ ~ f(x) = \frac{\frac{1}{x} + \frac{1}{2}}{x^3 + 8} ~ \]

Find the limit as $x$ goes to $27$ of

\[ ~ f(x) = \frac{x - 27}{x^{1/3} - 3} ~ \]

Find the limit

\[ ~ L = \lim_{x \rightarrow \pi/2} \frac{\tan (2x)}{x - \pi/2} ~ \]

The limit of $\sin(x)/x$ at $0$ has a numeric value. This depends upon the fact that $x$ is measured in radians. Try to find this limit: `limit(sind(x)/x, x=>0)`

. What is the value?

What is the limit `limit(sinpi(x)/x, x=>0)`

?

There are several properties of limits that allow one to break down more complicated problems into smaller subproblems. For example,

\[ ~ \lim (f(x) + g(x)) = \lim f(x) + \lim g(x) ~ \]

is notation to indicate that one can take a limit of the sum of two function or take the limit of each first, then add and the answer will be unchanged, provided all the limits in question exist.

Use one or the either to find the limit of $f(x) = \sin(x) + \tan(x) + \cos(x)$ as $x$ goes to $0$.

Does this function have a limit as $h$ goes to $0$ from the right (that is, assume $h>0$)?

\[ ~ \frac{h^h - 1}{h} ~ \]

Compute the limit

\[ ~ \lim_{x \rightarrow 1} \frac{x}{x-1} - \frac{1}{\log(x)}. ~ \]

Compute the limit

\[ ~ \lim_{x \rightarrow 1/2} \frac{1}{\pi} \frac{\cos(\pi x)}{1 - (2x)^2}. ~ \]