We use derivatives to give us a “short-cut” for computing limits.
Derivatives allow us to take problems that were once difficult to solve and
convert them to problems that are easier to solve. Let us consider L’Hôpital’s
rule:
L’Hôpital’s Rule Let
\(f(x)\) and
\(g(x)\) be functions that are differentiable near
\(a\) . If
\[ \lim _{x \to a} f(x) = \lim _{x \to a}g(x) = 0 \qquad \text {or} \pm \infty , \]
and
\(g'(x) \neq 0\) for all
\(x\)
near
\(a\) , then
\[ \lim _{x \to a} \frac {f(x)}{g(x)} = \lim _{x \to a} \frac {f'(x)}{g'(x)}, \]
provided that
\(\lim _{x \to a} \frac {f'(x)}{g'(x)}\) exists or
\(\pm \infty \) .
This theorem is somewhat difficult to prove, in part because it incorporates so many
different possibilities, so we will not prove it here.
L’Hôpital’s rule allows us to investigate limits of indeterminate form .
List of Indeterminate Forms
\(\relax \boldsymbol {\tfrac {0}{0}}\)
This refers to a limit of the form \(\lim _{x\to a} \frac {f(x)}{g(x)}\) where \(f(x)\to 0\) and \(g(x)\to 0\) as \(x\to a\) .
\(\relax \boldsymbol {\tfrac {\infty }{\infty }}\)
This refers to a limit of the form \(\lim _{x\to a} \frac {f(x)}{g(x)}\) where \(f(x)\to \infty \) and \(g(x)\to \infty \) as \(x\to a\) .
\(\relax \small \boldsymbol {0\cdot \infty }\)
This refers to a limit of the form \(\lim _{x\to a} \left (f(x)\cdot g(x)\right )\) where \(f(x)\to 0\) and \(g(x)\to \infty \) as \(x\to a\) .
\(\relax \small \boldsymbol {\infty - \infty }\)
This refers to a limit of the form \(\lim _{x\to a}\left ( f(x)-g(x)\right )\) where \(f(x)\to \infty \) and \(g(x)\to \infty \) as \(x\to a\) .
\(\relax \boldsymbol {1^\infty }\)
This refers to a limit of the form \(\lim _{x\to a} f(x)^{g(x)}\) where \(f(x)\to 1\) and \(g(x)\to \infty \) as \(x\to a\) .
\(\relax \boldsymbol {0^0}\)
This refers to a limit of the form \(\lim _{x\to a} f(x)^{g(x)}\) where \(f(x)\to 0\) and \(g(x)\to 0\) as \(x\to a\) .
\(\relax \boldsymbol {\infty ^0}\)
This refers to a limit of the form \(\lim _{x\to a} f(x)^{g(x)}\) where \(f(x)\to \infty \) and \(g(x)\to 0\) as \(x\to a\) .
In each of these cases, the value of the limit is not immediately obvious. Hence, a
careful analysis is required!
Our first example is the computation of a limit that was somewhat difficult
before.
Compute
\[ \lim _{x\to 0} \frac {\sin (x)}{x}. \]
Set
\(f(x) = \sin (x)\) and
\(g(x) = x\) . Since both
\(f(x)\) and
\(g(x)\) are differentiable functions at
\(0\) , and
\[ \lim _{x \to 0} f(x) = \lim _{x \to 0}g(x) = 0, \]
this
situation is ripe for L’Hôpital’s Rule. Now
\[ f'(x) = \answer [given]{\cos (x)} \]
and
\[ g'(x) = \answer [given]{1}. \]
L’Hôpital’s rule tells us that
\[ \lim _{x \to 0} \frac {\sin (x)}{x} = \lim _{x \to 0} \frac {\cos (x)}{1} = 1. \]
Compute:
\[ \lim _{x\to \infty } \dfrac {3x^2 + 1}{2xe^x}.\]
Set
\(f(x) =3x^2+1\) and
\(g(x)=2xe^x\) . Since
\(\displaystyle \lim _{x\to \infty }f(x) = \infty \) and
\(\displaystyle \lim _{x\to \infty } g(x) = \infty \) this is a limit with form
\(\inftyOverInfty \) , so L’Hôpital’s Rule (LHR) applies.
This gives:
\[ \lim _{x\to \infty } \dfrac {3x^2 + 1}{2xe^x} = \lim _{x\to \infty }\dfrac {6x}{2e^x + 2xe^x}. \]
This right-hand side is another limit with form
\(\inftyOverInfty \) , so we can use LHR
again.
\[ \lim _{x\to \infty }\dfrac {6x}{2e^x + 2xe^x} = \lim _{x\to \infty }\dfrac {6}{4e^x + 2xe^x}. \]
This limit has numerator tending to
\(6\) and denominator tending to
\(\infty \) , making
this limit have value
\(0\) .
\[ \lim _{x\to \infty }\dfrac {3x^2+1}{2xe^x} = 0\]
Our next set of examples will run through the remaining indeterminate forms one is
likely to encounter.
\(\zeroTimesInfty \) forms arise from a limit of the form \(\displaystyle \lim _{x\to a}f(x)g(x)\) . One way to write \(f(x)g(x)\) as a fraction remember that
\[ f(x)g(x) = \dfrac {f(x)}{\left (\dfrac {1}{g(x)}\right )} = \dfrac {g(x)}{\left (\dfrac {1}{f(x)}\right )}. \]
Compute
\[ \lim _{x\to \pi /2^+} \frac {\sec (x)}{\tan (x)}. \]
Set
\(f(x) = \sec (x)\) and
\(g(x) = \tan (x)\) . Both
\(f(x)\) and
\(g(x)\) are differentiable near
\(\pi /2\) . Additionally,
\[ \lim _{x \to \pi /2^+} f(x) = \lim _{x \to \pi /2^+}g(x) = -\infty . \]
This
situation is ripe for L’Hôpital’s Rule. Now
\[ f'(x) = \answer [given]{\sec (x)\tan (x)} \]
and
\[ g'(x) = \answer [given]{\sec ^2(x)}. \]
L’Hôpital’s rule tells us that
\begin{align*} \lim _{x\to \pi /2^+} \frac {\sec (x)}{\tan (x)} &= \lim _{x\to \pi /2^+} \frac {\sec (x)\tan (x)}{\sec ^2(x)} \\ &= \lim _{x\to \pi /2^+} \sin (x)\\ &=\answer [given]{1}. \end{align*}
Compute
\[ \lim _{x\to 0^+} x\ln (x). \]
This doesn’t appear to be suitable for L’Hôpital’s Rule. As
\(x\) approaches
zero,
\(\ln (x)\) goes to
\(-\infty \) , so the product looks like
\[ (\text {something very small})\cdot (\text {something very large and negative}). \]
This product could be anything.
A careful analysis is required. Write
\[ x\ln (x) = \frac {\ln (x)}{x^{-1}}. \]
Set
\(f(x) = \ln (x)\) and
\(g(x) = x^{-1}\) . Since both functions are
differentiable near zero and
\[ \lim _{x\to 0+} \ln (x) = -\infty \qquad \text {and}\qquad \lim _{x\to 0+} x^{-1} = \infty , \]
we may apply L’Hôpital’s rule. Write with me
\[ f'(x) = \answer [given]{x^{-1}} \]
and
\[ g'(x) = \answer [given]{-x^{-2}}, \]
so
\begin{align*} \lim _{x\to 0^+} x\ln (x) &= \lim _{x\to 0^+} \frac {\ln (x)}{x^{-1}} \\ &= \lim _{x\to 0^+} \frac {x^{-1}}{-x^{-2}}\\ &=\lim _{x\to 0^+} -x \\ &= 0. \end{align*}
One way to interpret this is that since \(\lim _{x\to 0^+}x\ln (x) = 0\) , the function \(x\) approaches zero much faster
than \(\ln (x)\) approaches \(-\infty \) .
There are two basic cases here, we’ll do an example of each.
Compute
\[ \lim _{x\to 0} \left (\cot (x) - \csc (x)\right ). \]
Here we simply need to write each term as a fraction,
\begin{align*} \lim _{x\to 0} \left (\cot (x) - \csc (x)\right ) &= \lim _{x\to 0} \left (\frac {\cos (x)}{\sin (x)} - \frac {1}{\sin (x)}\right )\\ &= \lim _{x\to 0} \frac {\cos (x)-1}{\sin (x)} \end{align*}
Setting \(f(x) = \cos (x)-1\) and \(g(x)=\sin (x)\) , both functions are differentiable near zero and
\[ \lim _{x\to 0}(\cos (x)-1)=\lim _{x\to 0}\sin (x) = 0. \]
We may now apply
L’Hôpital’s rule. Write with me
\[ f'(x) = \answer [given]{-\sin (x)} \]
and
\[ g'(x) = \answer [given]{\cos (x)}, \]
so
\begin{align*} \lim _{x\to 0} \left (\cot (x) - \csc (x)\right ) &= \lim _{x\to 0} \frac {\cos (x)-1}{\sin (x)}\\ &= \lim _{x\to 0} \frac {-\sin (x)}{\cos (x)} \\ &=0. \end{align*}
Sometimes one must be slightly more clever.
Compute
\[ \lim _{x\to \infty }\left (\sqrt {x^2+x}-x\right ). \]
Again, this doesn’t appear to be suitable for L’Hôpital’s Rule. A bit of
algebraic manipulation will help. Write with me
\begin{align*} \lim _{x\to \infty }\left (\sqrt {x^2+x}-x\right ) &= \lim _{x\to \infty }\left (x\left (\sqrt {1+1/x}-1\right )\right )\\ &=\lim _{x\to \infty }\frac {\sqrt {1+1/x}-1}{x^{-1}} \end{align*}
Now set \(f(x) = \sqrt {1+1/x}-1\) , \(g(x) = x^{-1}\) . Since both functions are differentiable for large values of \(x\) and
\[ \lim _{x\to \infty } (\sqrt {1+1/x}-1) = \lim _{x\to \infty }x^{-1} = 0, \]
we may
apply L’Hôpital’s rule. Write with me
\[ f'(x) = \answer [given]{(1/2)(1+1/x)^{-1/2}\cdot (-x^{-2})} \]
and
\[ g'(x) = \answer [given]{-x^{-2}} \]
so
\begin{align*} \lim _{x\to \infty }\left (\sqrt {x^2+x}-x\right ) &= \lim _{x\to \infty }\frac {\sqrt {1+1/x}-1}{x^{-1}} \\ &= \lim _{x\to \infty }\frac {(1/2)(1+1/x)^{-1/2}\cdot (-x^{-2})}{-x^{-2}} \\ &= \lim _{x\to \infty } \frac {1}{2\sqrt {1+1/x}}\\ &= \frac {1}{2}. \end{align*}
There is a standard trick for dealing with the indeterminate forms
\[ \oneToInfty ,\quad \zeroToZero ,\quad \inftyToZero . \]
Given \(u(x)\) and \(v(x)\) such
that
\[ \lim _{x\to a}u(x)^{v(x)} \]
falls into one of the categories described above, rewrite as
\[ \lim _{x\to a}e^{\ln \left (u(x)^{v(x)}\right )} \]
and then we rewrite
as
\[ \lim _{x\to a}e^{v(x)\ln {(u(x))}}. \]
Recall that the exponential function is continuous. Therefore
\[ \lim _{x\to a}e^{v(x)\ln {(u(x))}}=e^{\left [\lim _{x\to a} v(x)\ln (u(x))\right ]}. \]
Now we will focus
on the limit
\[ \lim _{x\to a} v(x)\ln (u(x)) \]
using L’Hôpital’s rule. (After calculating this limit, don’t forget to plug
back into the exponential.)
In the last two examples below, notice that they are different exponential
indeterminate forms, but we approach them with the same technique.
First determine
the form of the limit, then compute the limit.
\[ \lim _{x\to 0^+} x ^{\sin (x)}. \]
Select the correct choice. The form of
the limit is
\(\zeroOverZero \) \(\relax \boldsymbol {\tfrac {\infty }{\infty }}\) \(\zeroTimesInfty \) \(\relax \small \boldsymbol {\infty - \infty }\) \(\oneToInfty \) \( \inftyToZero \) \(\zeroToZero \)
Write
\[\lim _{x\to 0^+} x^{\sin (x)} = e^{\left [ \lim _{x\to 0^+} \sin (x) \ln (x) \right ]}.\]
Now we look at the limit in the exponent.
\[ \lim _{x\to 0^+} \sin (x) \ln (x) = \lim _{x\to 0^+} \dfrac {\ln (x)}{(\sin (x))^{-1}} =\lim _{x\to 0^+} \dfrac {\ln (x)}{\csc (x)}\]
Since
\(\displaystyle \lim _{x\to 0^+} \ln (x) = \answer [given]{-\infty }\) and
\(\displaystyle \lim _{x\to 0^+}\csc (x) = \answer [given]{\infty }\) , we know that
\(\displaystyle \lim _{x\to 0^+} \dfrac {\ln (x)}{\csc (x)}\) has
form
\(\inftyOverInfty \) , so we can apply l’Hôpital’s Rule.
That gives:
\begin{align*} \lim _{x\to 0^+} \dfrac {\ln (x)}{\csc (x)} &= \lim _{x\to 0^+} \dfrac { \frac {1}{x} }{-\csc (x)\cot (x)}, \textrm { l'H\^{o}pital's Rule}\\ &= \lim _{x\to 0^+} \dfrac {- \sin (x)\tan (x)}{x}, \textrm { algebra}\\ &= \lim _{x\to 0^+} \dfrac {-\sin (x)\sec ^2(x) - \cos (x)\tan (x)}{1}, \textrm {l'H\^{o}pital's Rule}\\ &= \dfrac {0 - 0}{1} = 0. \end{align*}
Notice that \(\displaystyle \lim _{x\to 0^+} \dfrac {- \sin (x)\tan (x)}{x}\) had form \(\zeroOverZero \) , we used l’Hôpital’s Rule again in that step.
Remember that \(\displaystyle \lim _{x\to 0^+} x^{\sin (x)} = e^{\left [ \lim _{x\to 0^+} \sin (x) \ln (x) \right ]}\) . We just calculated that \(\displaystyle \lim _{x\to 0^+} \sin (x)\ln (x) = 0\) , which means:
\[\lim _{x\to 0^+} x^{\sin (x)} = \answer [given]{1}\]
First determine the form of the limit, then compute the limit.
\[ \lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x. \]
Select the correct
choice. The form of the limit is
\(\zeroOverZero \) \(\relax \boldsymbol {\tfrac {\infty }{\infty }}\) \(\zeroTimesInfty \) \(\relax \small \boldsymbol {\infty - \infty }\) \(\oneToInfty \) \( \inftyToZero \) \(\zeroToZero \)
Write
\[ \lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x = e^{\left [\lim _{x\to \infty }x\ln \left (1 + \frac {1}{x}\right )\right ]}. \]
So now look at the
limit of the exponent
\[ \lim _{x\to \infty } x\ln \left (1 + \frac {1}{x}\right ) = \lim _{x\to \infty } \frac {\ln \left (1 + \frac {1}{x}\right )}{x^{-1}}. \]
Setting
\(f(x) = \ln \left (1 + \frac {1}{x}\right )\) and
\(g(x) = x^{-1}\) , both functions are differentiable for
large values of
\(x\) and
\[ \lim _{x\to \infty }\ln \left (1 + \frac {1}{x}\right )=\lim _{x\to \infty }x^{-1} = 0. \]
We may now apply L’Hôpital’s rule. Write
\[ f'(x) = \answer [given]{\frac {-x^{-2}}{1 + \frac {1}{x}}} \]
and
\[ g'(x) = \answer [given]{-x^{-2}}, \]
so
\begin{align*} \lim _{x\to \infty } \frac {\ln \left (1 + \frac {1}{x}\right )}{x^{-1}} &= \lim _{x\to \infty } \frac {\frac {-x^{-2}}{1 + \frac {1}{x}}}{-x^{-2}} \\ &=\lim _{x\to \infty } \frac {1}{1 + \frac {1}{x}}\\ &=1. \end{align*}
Hence,
\begin{align*} \lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x &=e^{ \eval {\lim _{x\to \infty }x\ln \left (1 + \frac {1}{x}\right )}} \\ &=e^{ \eval {\lim _{x\to \infty } \frac {\ln \left (1 + \frac {1}{x}\right )}{x^{-1}}}} \\ &=e^{1} \\ &= e. \end{align*}