5.17 L'Hôpital's RuleHL ONLY

A First Discussion

We know that $\dfrac{0}{5} = 0$ and $\dfrac{5}{0}$ is not defined. However, when $x$ tends to 0, $\dfrac{5}{x}$ tends either to $+\infty$ or $-\infty$. But what about $\dfrac{0}{0}$?

Consider a function of the form $\dfrac{f(x)}{g(x)}$

  • If $f(x) \to 0$ and $g(x) \to 5$ then $\dfrac{f(x)}{g(x)}$ tends to 0.
    e.g., $\lim_{x\to 5} \dfrac{x-5}{5} = 0$
  • If $f(x) \to 5$ and $g(x) \to 0$ then $\dfrac{f(x)}{g(x)}$ tends to $+\infty$ or $-\infty$.
    e.g., $\lim_{x\to 5} \dfrac{5}{x-5} = \pm\infty$

But again, what happens when both $f(x) \to 0$ and $g(x) \to 0$? The result could be anything: 0, $+\infty$ or $-\infty$ or any real number

For example, when $x \to 0$, all the functions below have the form $\dfrac{0}{0}$, however

$$ \lim_{x\to 0} \dfrac{3x}{x} = 3, \quad \lim_{x\to 0} \dfrac{x^3}{x} = 0, \quad \lim_{x\to 0} \dfrac{\sin x}{x} = 1, \quad \lim_{x\to 0} \dfrac{\sin(2x)}{x} = 2 $$
That is why we say that: $\dfrac{0}{0}$ is an indeterminate form.
In the same sense: $\dfrac{\infty}{\infty}$ is also an indeterminate form.

1. $\lim \frac{f(x)}{g(x)}$ In General

Here we deal with limits of the form $\lim_{x\to a} \dfrac{f(x)}{g(x)}$, $\lim_{x\to +\infty} \dfrac{f(x)}{g(x)}$, $\lim_{x\to -\infty} \dfrac{f(x)}{g(x)}$.
If the fraction is strictly not of the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ we can easily find the result.

For example,

$\displaystyle\lim_{x\to 1} \dfrac{2x+3}{3x+5} = \dfrac{5}{8}$ $\displaystyle\lim_{x\to 3} \dfrac{x-3}{x^2+9} = 0$ $\displaystyle\lim_{x\to +\infty} \dfrac{2}{3x+5} = 0$
$\displaystyle\lim_{x\to +\infty} \dfrac{e^{-x}}{3x} = 0$ $\displaystyle\lim_{x\to +\infty} \dfrac{x^3}{2x+7} = +\infty$ $\displaystyle\lim_{x\to -\infty} \dfrac{2x+7}{e^x} = -\infty$
$\displaystyle\lim_{x\to 3} \dfrac{2}{(x-3)^2} = +\infty$ $\displaystyle\lim_{x\to 3^-} \dfrac{2}{x-3} = -\infty$ $\displaystyle\lim_{x\to 3^+} \dfrac{2}{x-3} = +\infty$

2. The Indeterminate Forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$

In some cases the answer is easy. For example $\dfrac{\infty}{\infty}$,

  • $\lim_{x\to +\infty} \dfrac{4x+7}{2x+3} = \dfrac{4}{2} = 2$
  • $\lim_{x\to +\infty} \dfrac{4x+7}{2x^3+3} = 0$
  • $\lim_{x\to +\infty} \dfrac{4x^3+7}{2x+3} = +\infty$

For $\dfrac{0}{0}$,

$$ \lim_{x\to 3} \dfrac{x^2-3x}{x-3} = \lim_{x\to 3} \dfrac{x(x-3)}{x-3} = \lim_{x\to 3} x = 3 $$

It is also known that $$ \lim_{x\to 0} \dfrac{\sin x}{x} = 1 $$

For more complicated cases the following theorem helps; we simply need the derivatives of $f(x)$ and $g(x)$.

L'Hôpital's Rule

If $f(x) \to 0$ and $g(x) \to 0$ OR $f(x) \to \pm\infty$ and $g(x) \to \pm\infty$, then:

$\lim \dfrac{f(x)}{g(x)} = \lim \dfrac{f'(x)}{g'(x)}$

provided that $\lim \dfrac{f'(x)}{g'(x)}$ exists.

EXAMPLE 1

$\lim_{x\to +\infty} \dfrac{4x^2+7}{2x^2+3} \stackrel{\left(\frac{\infty}{\infty}\right)}{=} \lim_{x\to +\infty} \dfrac{8x}{4x} = \mathbf{2}$
Remember to appropriately indicate the form above the equals sign: $\stackrel{\left(\frac{0}{0}\right)}{=}$ or $\stackrel{\left(\frac{\infty}{\infty}\right)}{=}$.

Sometimes we need to apply L'Hôpital's rule more than once.

EXAMPLE 2

$\lim_{x\to +\infty} \dfrac{4x^2-7x+5}{x^3-2x} \stackrel{\left(\frac{\infty}{\infty}\right)}{=} \lim_{x\to +\infty} \dfrac{8x-7}{3x^2-2} \stackrel{\left(\frac{\infty}{\infty}\right)}{=} \lim_{x\to +\infty} \dfrac{8}{6x} = \mathbf{0}$

EXAMPLE 3

$\lim_{x\to 0} \dfrac{e^x-x-1}{x^2} \stackrel{\left(\frac{0}{0}\right)}{=} \lim_{x\to 0} \dfrac{e^x-1}{2x} \stackrel{\left(\frac{0}{0}\right)}{=} \lim_{x\to 0} \dfrac{e^x}{2} = \mathbf{\dfrac{1}{2}}$

EXAMPLE 4

Find the horizontal asymptotes of $f(x) = \dfrac{3e^{2x}+2}{e^{2x}-1}$

  • When $x \to +\infty$:
    $\lim_{x\to +\infty} \dfrac{3e^{2x}+2}{e^{2x}-1} \stackrel{\left(\frac{\infty}{\infty}\right)}{=} \lim_{x\to +\infty} \dfrac{6e^{2x}}{2e^{2x}} = 3$
    H.A. is $\mathbf{y=3}$.
  • When $x \to -\infty$:
    The limit is not of the form $\dfrac{\infty}{\infty}$ since $e^{2x} \to 0$.
    $\lim_{x\to -\infty} \dfrac{3e^{2x}+2}{e^{2x}-1} = \dfrac{2}{-1} = -2$
    H.A. is $\mathbf{y=-2}$.

3. Other Inderterminate Forms

The following are also indeterminate forms: $\infty \cdot 0$ , $\infty - \infty$ , $1^\infty$.

For example, look at the following $\infty \cdot 0$ forms:

  • $\lim_{x\to 0} x \cdot \left(\dfrac{1}{x}\right) = \lim_{x\to 0} 1 = 1$
  • $\lim_{x\to 0} x^2 \cdot \left(\dfrac{1}{x}\right) = \lim_{x\to 0} x = 0$
  • $\lim_{x\to 0} x \cdot \left(\dfrac{1}{x^2}\right) = \lim_{x\to 0} \dfrac{1}{x} = +\infty$

More complicated forms can be mathematically transformed to the absolute form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ and thus answered by utilizing L'Hôpital's rule.

EXAMPLE 5

When $x \to 0$ then $\ln x \to -\infty$. Thus $\lim_{x\to 0} x \ln x$ is of the form $0 \cdot \infty$. But

$\lim_{x\to 0} x \ln x = \lim_{x\to 0} \dfrac{\ln x}{\dfrac{1}{x}} \stackrel{\left(\frac{-\infty}{+\infty}\right)}{=} \lim_{x\to 0} \dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^2}} = \lim_{x\to 0} (-x) = \mathbf{0}$

EXAMPLE 6

The limit $\lim_{x\to +\infty} \left(x - \sqrt{x^2+1}\right)$ is uniquely of the form $\infty - \infty$. But:

$$ \begin{aligned} \lim_{x\to +\infty} \left(x - \sqrt{x^2+1}\right) &= \lim_{x\to +\infty} \dfrac{\left(x - \sqrt{x^2+1}\right)\left(x + \sqrt{x^2+1}\right)}{x + \sqrt{x^2+1}} \\ &= \lim_{x\to +\infty} \dfrac{x^2 - (x^2+1)}{x + \sqrt{x^2+1}} \\ &= \lim_{x\to +\infty} \dfrac{-1}{x + \sqrt{x^2+1}} \\ &= \mathbf{0} \end{aligned} $$

EXAMPLE 7

We will structurally demonstrate that $\lim_{x\to +\infty} \left(1 + \dfrac{1}{x}\right)^x = e$. It holds the indeterminate form $1^\infty$. But,

$$ \left(1 + \dfrac{1}{x}\right)^x = e^{\ln\left(1 + \frac{1}{x}\right)^x} = e^{x \ln\left(1 + \frac{1}{x}\right)} $$ Let's isolate and evaluate the absolute limit of the exponent $x \ln\left(1 + \dfrac{1}{x}\right)$; it takes the functional form $\infty \cdot 0$:
$$ \begin{aligned} \lim_{x\to +\infty} x \ln\left(1 + \dfrac{1}{x}\right) &= \lim_{x\to +\infty} \dfrac{\ln\left(1 + \dfrac{1}{x}\right)}{\dfrac{1}{x}} \\ &\stackrel{\left(\frac{0}{0}\right)}{=} \lim_{x\to +\infty} \dfrac{\dfrac{1}{1+\dfrac{1}{x}} \cdot \left(-\dfrac{1}{x^2}\right)}{-\dfrac{1}{x^2}} \\ &= \lim_{x\to +\infty} \dfrac{1}{1+\dfrac{1}{x}} \\ &= \mathbf{1} \end{aligned} $$ Hence, $$ \lim_{x\to +\infty} \left(1 + \dfrac{1}{x}\right)^x = \lim_{x\to +\infty} e^{x \ln\left(1 + \frac{1}{x}\right)} = e^1 = \mathbf{e} $$
NOTICE:
  • In a precisely similar way we can show that $\lim_{x\to +\infty} \left(1 + \dfrac{a}{x}\right)^x = e^a$.
  • As we said earlier we accept the limit $\lim_{x\to 0} \dfrac{\sin x}{x} = 1$ as known. Although it is strictly of the form $\dfrac{0}{0}$ and the rule would inherently give:
    $\lim_{x\to 0} \dfrac{\sin x}{x} \stackrel{\left(\frac{0}{0}\right)}{=} \lim_{x\to 0} \dfrac{\cos x}{1} = 1$
    we mathematically cannot use L'Hôpital! This is strictly because the proof of the fact $(\sin x)' = \cos x$ (by first principles) already natively uses the limit $\lim_{x\to 0} \dfrac{\sin x}{x} = 1$. The proof of this limit in particular uses other profound trigonometric techniques and it is fully beyond the scope of this course.