L'Hospital's rule is the definitive way to simplify evaluation of
limits. It does not directly evaluate limits, but only *simplifies
evaluation if used appropriately*.

In effect, this rule is the ultimate version of ‘cancellation tricks’, applicable in situations where a more down-to-earth genuine algebraic cancellation may be hidden or invisible.

Suppose we want to evaluate
$$\lim_{x\rightarrow a}{f(x)\over g(x)}$$
where the limit $a$ could also be $+\infty$ or $-\infty$ in addition
to ‘ordinary’ numbers. Suppose that *either*
$$\lim_{x\rightarrow a}f(x)=0 \hbox{ and } \lim_{x\rightarrow a}
g(x)=0$$
*or*
$$\lim_{x\rightarrow a}f(x)=\pm\infty \hbox{ and } \lim_{x\rightarrow a}
g(x)=\pm\infty$$
(The $\pm$'s don't have to be the same sign). Then we cannot just
‘plug in’ to evaluate the limit, and these are traditionally called
**indeterminate forms**. The unexpected trick that works often
is that (amazingly) we are entitled to *take the derivative of
both numerator and denominator:*
$$\lim_{x\rightarrow a}{f(x)\over g(x)}=
\lim_{x\rightarrow a}{f'(x)\over g'(x)}.$$
No, this is *not the quotient rule*. No, it is not so clear why
this would help, either, but we'll see in examples.

#### Example 1

Find $\lim_{x\rightarrow 0}\,(\sin\,x)/x$.

**Solution:** both numerator and
denominator have limit $0$, so we are entitled to apply L'Hospital's
rule:
$$\lim_{x\rightarrow 0}\,{\sin\,x\over x}=
\lim_{x\rightarrow 0}\,{\cos\,x\over 1}.$$
In the new expression, *neither* numerator nor denominator is $0$
at $x=0$, and we can just plug in to see that the limit is $1$.

#### Example 2

Find $\lim_{x\rightarrow 0}\,x/(e^{2x}-1)$.

**Solution:** both numerator
and denominator go to $0$, so we are entitled to use L'Hospital's
rule:
$$\lim_{x\rightarrow 0}\,{ x \over e^{2x }-1}=
\lim_{x\rightarrow 0}\,{ 1 \over 2e^{2x }}$$
In the new expression, the numerator and denominator are both non-zero
when $x=0$, so we just plug in $0$ to get
$$\lim_{x\rightarrow 0}\,{ x \over e^{2x }-1}=
\lim_{x\rightarrow 0}\,{ 1 \over 2e^{2x }}={1\over 2e^0}={1\over 2}$$

#### Example 3

Find $\lim_{x\rightarrow 0^+}\,x\,\ln x$.

**Solution:** The $0^+$ means
that we approach $0$ from the positive side, since otherwise we won't
have a real-valued logarithm. This problem
illustrates the *possibility* as well as *necessity* of *rearranging* a limit to make it be a *ratio* of things, in order
to legitimately apply L'Hospital's rule. Here, we rearrange to
$$\lim_{x\rightarrow 0^+}\,x\,\ln x=\lim_{x\rightarrow
0}{\ln\,x\over 1/x}$$
In the new expressions the top goes to $-\infty$ and the bottom goes
to $+\infty$ as $x$ goes to $0$ (from the right). Thus, we are
entitled to apply L'Hospital's rule, obtaining
$$\lim_{x\rightarrow 0^+}\,x\,\ln x=\lim_{x\rightarrow
0}{\ln\,x\over 1/x}=\lim_{x\rightarrow
0}{1/x\over -1/x^2}$$
Now it is very necessary to rearrange the expression inside the last
limit: we have
$$\lim_{x\rightarrow 0}{1/x\over -1/x^2}=
\lim_{x\rightarrow 0}\,-x$$
The new expression is very easy to evaluate: the limit is $0$.

#### Example 4

It is often necessary to apply L'Hospital's rule repeatedly: Let's find $\lim_{x\rightarrow +\infty}x^2/e^x$.

**Solution:** both numerator and
denominator go to $\infty$ as $x\rightarrow +\infty$, so we are
entitled to apply L'Hospital's rule, to turn this into
$$\lim_{x\rightarrow +\infty}{2x\over e^x}$$
But still both numerator and denominator go to $\infty$, so apply
L'Hospital's rule again: the limit is
$$\lim_{x\rightarrow +\infty}{2\over e^x}=0$$
since now the numerator is fixed while the denominator goes to $+\infty$.

#### Example 5

Now let's illustrate more ways that things can be rewritten as ratios, thereby possibly making L'Hospital's rule applicable. Let's evaluate $$\lim_{x \rightarrow 0}\,x^x.$$

**Solution:**
It is less obvious now, but we can't just plug in $0$ for $x$: on one
hand, we are taught to think that $x^0=1$, but also that $0^x=0$; but
then surely $0^0$ can't be both at once. And this exponential
expression is not a ratio.

The trick here is to *take the logarithm*:
$$\ln\left(\lim_{x \rightarrow 0^+}\,x^x\right)=\lim_{x \rightarrow 0^+}\,\ln(x^x)$$
The reason that we are entitled to *interchange* the logarithm and
the limit is that *logarithm is a continuous function* (on its
domain). Now we use the fact that $\ln(a^b)=b\ln a$, so the log of the
limit is
$$\lim_{x \rightarrow 0^+}\,x\ln x$$
Aha! The question has been turned into one we already did! But
ignoring that, and repeating ourselves, we'd first rewrite this as a
ratio
$$\lim_{x \rightarrow 0^+}\,x\ln x=\lim_{x \rightarrow 0^+}\,{\ln
x\over 1/x}$$
and then apply L'Hospital's rule to obtain
$$\lim_{x \rightarrow 0^+}\,{1/x\over -1/x^2}=\lim_{x \rightarrow
0^+}\,-x=0$$
But we have to remember that we've computed the *log* of the
limit, not the limit. Therefore, the actual limit is
$$\lim_{x \rightarrow 0^+}\,x^x=e^{\hbox{ log of the limit
}}=e^0=1$$
*This trick of taking a logarithm is important to remember*.

#### Example 6

Here is another issue of rearranging to fit into accessible form: Find $$\lim_{x\rightarrow +\infty}\sqrt{x^2+x+1}-\sqrt{x^2+1}.$$

**Solution:**
This is not a ratio, but certainly is ‘indeterminate’, since it is the
difference of two expressions both of which go to $+\infty$. To make
it into a ratio, we take out the largest reasonable power of $x$:
\begin{align*}
\lim_{x\rightarrow +\infty}
\sqrt{x^2+x+1}-\sqrt{x^2+1}&=
\lim_{x\rightarrow +\infty} x\cdot\left(\sqrt{1+{1\over x}+{1\over x^2}}-
\sqrt{1+{1\over x^2}}\right)
\\
&=\lim_{x\rightarrow +\infty}
{\sqrt{1+{1\over x}+{1\over x^2}}-
\sqrt{1+{1\over x^2}} \over 1/x}
\end{align*}
The last expression here fits the requirements of the L'Hospital rule,
since both numerator and denominator go to $0$. Thus, by invoking
L'Hospital's rule, it becomes
$$=\lim_{x\rightarrow +\infty}
\;{1\over 2}\;
{{-{1\over x^2}-{2\over x^3} \over
\sqrt{1+{1\over x}+{1\over x^2}}}-
{ {-2\over x^3} \over \sqrt{1+{1\over x^2}}} \over -1/x^2}.$$

This is a large but actually tractable expression: multiply top and bottom by $x^2$, so that it becomes $$=\lim_{x\rightarrow +\infty} {{1\over 2}+{1\over x} \over \sqrt{1+{1\over x}+{1\over x^2}}}+ { {-1\over x} \over \sqrt{1+{1\over x^2}} } .$$

At this point, we *can* replace every ${1\over x}$ by $0$,
finding that the limit is equal to
$${{1\over 2}+0 \over
\sqrt{1+0+0}}+
{ 0 \over \sqrt{1+0 }}={1\over 2}.$$

It is important to recognize that in additional to the actual
application of L'Hospital's rule, it may be necessary to *experiment* a little to get things to settle out the way you
want. *Trial-and-error is not only ok, it is necessary*.

#### Exercises

- Find $\lim_{x\rightarrow 0}\,(\sin\,x)/x$
- Find $\lim_{x\rightarrow 0}\,(\sin\,5x)/x$
- Find $\lim_{x\rightarrow 0}\,(\sin\,(x^2))/x^2$
- Find $\lim_{x\rightarrow 0}\,x/(e^{2x}-1)$
- Find $\lim_{x\rightarrow 0}\,x\,\ln x$
- Find $$\lim_{x\rightarrow 0^+}\,(e^x-1)\ln x$$
- Find $$\lim_{x\rightarrow 1}\,{ \ln x \over x-1 }$$
- Find $$\lim_{x\rightarrow +\infty}\,{ \ln\,x \over x }$$
- Find $$\lim_{x\rightarrow +\infty}\,{ \ln x \over x^2 }$$
- Find $\lim_{x\rightarrow 0}\,(\sin x)^x$