6 L'Hôpital's Rule and indeterminate forms

Try making one function a multiple of the other.

There are many possible answers. Here is one: $f(x)=5x$, $g(x)=2x$.

There are many possible answers. Consider these: $f(x)=5x$, $g(x)=2x$. Then $\ds\lim_{x \rightarrow\infty}f(x)=\ds\lim_{x\to\infty}g(x)=\infty$, and $\ds\lim_{x\to\infty}\frac{f(x)}{g(x)}=\ds\lim_{x\to\infty}\frac{5x}{2x}=\ds\lim_{x\to\infty}\frac{5}{2}=\frac{5}{2}=2.5$.

Try making one function a multiple of the other, but not a constant multiple.

There are many possible answers. Here is one: $f(x)=x$, $g(x)=x^2$.

There are many possible answers. Consider these: $f(x)=x$, $g(x)=x^2$. Then $\ds\lim_{x \rightarrow\infty}f(x)=\ds\lim_{x\to\infty}g(x)=\infty$, and $\ds\lim_{x\to\infty}\frac{f(x)}{g(x)}=\ds\lim_{x\to\infty}\frac{x}{x^2}=\ds\lim_{x\to\infty}\frac{1}{x}=0$.

Plugging in $x=1$ to the numerator and denominator makes both zero. This is exactly one of the indeterminate forms where l'H^opital's rule can be directly applied.

$-\dfrac{2}{\pi}$

If we plug in $x=1$ to the numerator and the denominator, we find they are both zero. So, we have an indeterminate form appropriate for L'H^opital's Rule.

Is this an indeterminate form?

$-\infty$

Be careful– this is not an indeterminate form! As $x\rightarrow 0+$, the numerator $\log x\rightarrow-\infty$. That is, the numerator is becoming an increasingly huge, negative number. As $x\rightarrow 0+$, the denominator $x\rightarrow 0+$, which only serves to make the total fraction even larger, and still negative. So, $\lim\limits_{x\rightarrow 0^+}\dfrac{\log x}{x} =-\infty$. Remark: if we had tried to use l'H^opital's Rule here, we would have come up with the wrong answer. If we differentiate the numerator and the denominator, the fraction becomes $\dfrac{\frac{1}{x}}{1}=\frac{1}{x}$, and $\ds\lim_{x \to 0^+}\frac{1}{x}=\infty$. The reason we cannot apply l'H^opital's Rule is that we do not have an indeterminate form, like both numerator and denominator going to infinity, or both numerator and denominator going to zero.

First, rearrange the expression to a more natural form (without a negative exponent).

0

We rearrange the expression to a more natural form:

Both the numerator and denominator go to infinity as $x$ goes to infinity. So, we can apply l'H^opital's Rule. In fact, we end up applying it twice.

The numerator gets smaller and smaller while the denominator gets larger and larger, so:

If at first you don't succeed, try, try again.

0

Keep at it!

3

Rather than use l'H^opital, try factoring out $x^2$ from the numerator and denominator.

$2$

If we plug in $x=0$ to the numerator and denominator, both are zero, so this is a candidate for l'H^opital's Rule. However, an easier way to evaluate the limit is to factor $x^2$ from the numerator and denominator, and cancel.

Keep going!

0

$\frac{1}{2}$

Try plugging in $x=0$. Is this an indeterminate form?

0

If we plug in $x=0$, the numerator is zero, and the denominator is
$\sec 0 = \dfrac{1}{\cos 0}=\dfrac{1}{1}=1$. So the limit is $\dfrac{0}{1}=0$.

Be careful: you cannot use l'H^opital's Rule here, because the fraction does not give an indeterminate form. If you try to differentiate the numerator and the denominator, you get an expression whose limit does not exist:
$\ds\lim_{x \to 0}\dfrac{1}{\sec x \tan x}=\ds\lim_{x \to 0}\cos x\cdot \dfrac{\cos x}{\sin x}=DNE$.

Simplify the trigonometric part first.

$5$

If we plug $x=0$ into the denominator, we get 1. However, the numerator is an indeterminate form: $\tan 0 =0$, while $\ds\lim_{x \to 0^+}\csc x=\infty$ and $\ds\lim_{x \to 0^-}\csc x=-\infty$. If we use $\csc x = \frac{1}{\sin x}$, our expression becomes

Since plugging in $x=0$ makes both the numerator and the denominator equal to zero, this is a candidate for l'H^ospital's Rule. However, a much easier way is to simplify the trig first.

If it is too difficult to take a derivative for l'H^opital's Rule, try splitting up the function into smaller chunks and evaluating their limits independently.

3

If we plug in $x=0$, both numerator and denominator become zero. So, we have exactly one of the indeterminate forms that l'H^opital's Rule applies to.

If we plug in $x=0$, still we find that both the numerator and the denominator go to zero. We could jump in with another iteration of l'H^opital's Rule. However, the derivatives would be a little messy, so we use limit laws and break up the fraction into the product of two fractions. If both limits exist:

We can evaluate the right-hand limit by simply plugging in $x=0$:

$\frac{3}{2}$

Try manipulating the function to get it into a nicer form

0

$\frac{1}{3}$

If the denominator tends to zero, and the limit exists, what must be the limit of the numerator?

$c=0$

Both the numerator and denominator converge to $0$ as $x\rightarrow 0$. So, by l'H^opital,

The new denominator still converges to $0$ as $x\rightarrow 0$. For the limit to exist, the same must be true for the new numerator. This tells us that if $c \neq 0$, the limit does not exist. We should check whether the limit exists when $c=0$. Using l'H^opital:

So, the limit exists when $c=0$.

Start with one application of l'H^opital's Rule. After that, you need to consider three distinct cases: $k>2$, $k<2$, and $k=2$.

$\lim\limits_{x\rightarrow 0}\dfrac{e^{k\sin(x^2)}-(1+2x^2)}{x^4}=\left\{\begin{array}{rl} -\infty&k<2\\ 2&k=2\\ \infty&k>2 \end{array}\right.$

The first thing we notice is, regardless of $k$, when we plug in $x=0$ both numerator and denominator become zero. Let's use this fact, and apply l'H^opital's Rule.

When we plug in $x=0$, the denominator becomes 0, and the numerator becomes $2k-4$. So, we'll need some cases, because the behaviour of the limit depends on $k$.

For $k=2$:

For $k>2$, the numerator goes to $2k-4$, which is a positive constant, while the denominator goes to $0$ from the right, so:

For $k<2$, the numerator goes to $2k-4$, which is a negative constant, while the denominator goes to $0$ from the right, so:

Percentage error: $100\left|\frac{\text{exact}-\text{approx}}{\text{exact}}\right|$. Absolute error: $|\text{exact}-\text{approx}|$.

Try modifying the function from Example 6.3.4.

There are many possible answers. Here is one: $f(x)=1+\frac{1}{x}$, $g(x)=x\log 5$ (recall we use $\log$ to mean logarithm base $e$).

From Example 6.3.4, we know that $\ds\lim_{x \to 0} (1+x)^{\frac{a}{x}}=e^a$, so $\ds\lim_{x \to 0} (1+x)^{\frac{\log 5}{x}}=e^{\log 5}=5$. However, this is the limit as $x$ goes to 0, which is not what we were asked. So, we modify the functions by replacing $x$ with $\frac{1}{x}$. If $x \to 0^+$, then $\frac{1}{x} \to \infty$.

Taking $f(x)=1+\frac{1}{x}$ and $g(x)=x\log 5$, we see:
(i) $\ds\lim_{x \to \infty} f(x)=\ds\lim_{x \to \infty } \left(1+\frac{1}{x}\right)=1$
(ii) $\ds\lim_{x \to \infty} g(x)=\ds\lim_{x \to \infty} x\log 5=\infty$
(iii) Let us name $\dfrac{1}{x}=X$. Then as $x \to \infty$, $X \to 0^+$, so:
$\ds\lim_{x \to \infty} [f(x)]^{g(x)}=\ds\lim_{x \to \infty} \left[1+\frac{1}{x}\right]^{x\log 5}=\lim_{x \to \infty} \left[1+\frac{1}{x}\right]^{\frac{\log 5}{\frac{1}{x}}}= \lim_{X \to 0^+} \left[1+X\right]^{\frac{\log 5}{X}}=e^{\log 5}=5$, where in the penultimate step, we used the result of Example 6.3.4.

$\lim\limits_{x \to 0}\sqrt[x^2]{\sin^2 x}=(\sin^2 x)^{\frac{1}{x^2}}$; what form is this?

0

$\ds\lim_{x \to 0} \sin^2 x = 0$, and $\ds\lim_{x \to 0}\frac{1}{x^2}=\infty$, so we have the form $0^\infty$. (Note that $\sin^2 x$ is positive, so our root is defined.) This is not an indeterminate form: $\lim\limits_{x \to 0}\sqrt[x^2]{\sin^2 x}=0$.

$\lim\limits_{x \to 0}\sqrt[x^2]{\cos x} = \ds\lim_{x \to 0}(\cos x)^{\frac{1}{x^2}}$

$\frac{1}{\sqrt{e}}$

$\ds\lim_{x \to 0} \cos x =1$ and $\ds\lim_{x \to 0}\frac{1}{x^2}=\infty$, so $\ds\lim_{x \to 0}(\cos x)^{\frac{1}{x^2}}$ has the indeterminate form $1^\infty$. We want to use l'H^opital, but we need to get our function into a fractional indeterminate form. So, we'll use a logarithm.

logarithms

1