Expansions and L'Hôpital's rule

What you will be able to do

• use the first few terms of $\sin x$, $\cos x$, $e^x$, $\ln(1 + x)$ and $\tan x$ to find a limit;
• state L'Hôpital's rule with its conditions and use it correctly;
• see when the rule fails and when a standard limit is safer.

Functions as short polynomials

Close to $x = 0$, a smooth function looks like a polynomial. The more terms we keep, the better the match. The standard expansions, for $x$ in radians, are

These are the Maclaurin series, which A-Level Further Maths and IB Higher Level prove. The Indian school syllabus does not prove them. JEE Main uses them as a tool, so we use them here. The standard limits of the first course are the first terms: $\dfrac{\sin x}{x} \to 1$ is the $x$ in the sine, and $\dfrac{e^x - 1}{x} \to 1$ is the $x$ in the exponential.

How to use them

1. Look at the top and the bottom. Find the lowest power of $x$ that survives after the terms cancel.
2. Replace each function by enough terms to see that power.
3. Divide, and let $x \to 0$.

Example 1. $\displaystyle\lim_{x \to 0}\dfrac{x - \sin x}{x^3}$. The top is $x - \left(x - \dfrac{x^3}{6} + \dots\right) = \dfrac{x^3}{6} - \dots$. Dividing by $x^3$ gives $\dfrac{1}{6}$.

Example 2. $\displaystyle\lim_{x \to 0}\dfrac{e^x - 1 - x}{x^2}$. The top is $\dfrac{x^2}{2} + \dfrac{x^3}{6} + \dots$. The limit is $\dfrac{1}{2}$.

Example 3. $\displaystyle\lim_{x \to 0}\dfrac{\tan x - x}{x^3} = \dfrac{1}{3}$, because the top is $\dfrac{x^3}{3} + \dots$.

Example 4, a $1^{\infty}$ form. $\displaystyle\lim_{x \to 0}(\cos x)^{1/x^2}$. Here $f - 1 = \cos x - 1 \approx -\dfrac{x^2}{2}$, so $g(f - 1) = \dfrac{1}{x^2} \cdot \left(-\dfrac{x^2}{2}\right) = -\dfrac{1}{2}$. The limit is $e^{-1/2}$.

L'Hôpital's rule

Statement. Let $f$ and $g$ be differentiable near $a$ (but not necessarily at $a$), with $g'(x) \ne 0$ near $a$. Suppose that

• $f(x) \to 0$ and $g(x) \to 0$ as $x \to a$, or $f(x) \to \pm\infty$ and $g(x) \to \pm\infty$; and
• $\displaystyle\lim_{x \to a}\dfrac{f'(x)}{g'(x)}$ exists, as a number or as $\pm\infty$.

Then $\displaystyle\lim_{x \to a}\dfrac{f(x)}{g(x)} = \lim_{x \to a}\dfrac{f'(x)}{g'(x)}$. The same holds for $x \to \pm\infty$.

The rule differentiates the top and the bottom separately. It is not the quotient rule. If $\dfrac{f'}{g'}$ is again $\dfrac{0}{0}$, use the rule again.

Example 5. $\displaystyle\lim_{x \to 0}\dfrac{e^x - 1 - x}{x^2}$. This is $\dfrac{0}{0}$. One step gives $\dfrac{e^x - 1}{2x}$, again $\dfrac{0}{0}$. A second step gives $\dfrac{e^x}{2} \to \dfrac{1}{2}$. This matches Example 2.

Example 6. $\displaystyle\lim_{x \to \infty}\dfrac{x^2}{e^x}$. This is $\dfrac{\infty}{\infty}$. Two steps give $\dfrac{2}{e^x} \to 0$. The exponential beats every power.

Example 7. $\displaystyle\lim_{x \to 0^+} x\ln x$. This is $0 \cdot \infty$. Write it as $\dfrac{\ln x}{1/x}$, which is $\dfrac{\infty}{\infty}$. The rule gives $\dfrac{1/x}{-1/x^2} = -x \to 0$. It follows that $x^x = e^{x\ln x} \to e^0 = 1$ as $x \to 0^+$.

Why the rule works

Look at two curves that both leave the origin. Very close to the origin, each one almost coincides with its tangent line, $f \approx f'(0)\,x$ and $g \approx g'(0)\,x$. So their ratio is almost $\dfrac{f'(0)}{g'(0)}$.

When the rule fails

Not an indeterminate form. For $\displaystyle\lim_{x \to 0}\dfrac{x + 1}{x + 2}$ the answer is $\dfrac{1}{2}$ by substitution. The rule would give $\dfrac{1}{1} = 1$, which is wrong. Check the form first.

The new limit does not exist. Take $\displaystyle\lim_{x \to \infty}\dfrac{2x + \sin x}{x + 3}$. The true limit is $2$, by dividing by $x$. The rule gives $\dfrac{2 + \cos x}{1}$, which has no limit, so the rule says nothing. The conclusion goes one way only: if $\dfrac{f'}{g'}$ has a limit, then $\dfrac{f}{g}$ has the same limit.

A good habit. In a school exam where L'Hôpital's rule is not allowed, use standard limits and algebra. In JEE and IB Higher Level it is a standard tool. Whichever you use, the answer must agree. You can check one method with the other.

Practise with new numbers

Try it in Python

The program adds the terms of the sine series and shows how fast it approaches $\sin 1$. Then it prints $\dfrac{x - \sin x}{x^3}$ for smaller and smaller $x$.

Common mistakes

• Using the rule when the form is not indeterminate. Substitute first.
• Applying the quotient rule. L'Hôpital differentiates the top and the bottom separately.
• Stopping the series too early. If the top becomes $0$ after the first terms, keep the next term.
• Trusting the rule when $\dfrac{f'}{g'}$ has no limit. Then the rule is silent. Try another method.

Remember these

• $\sin x \approx x - \dfrac{x^3}{6}$, $\cos x \approx 1 - \dfrac{x^2}{2}$, $e^x \approx 1 + x + \dfrac{x^2}{2}$, $\ln(1 + x) \approx x - \dfrac{x^2}{2}$, $\tan x \approx x + \dfrac{x^3}{3}$.
• L'Hôpital: for $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, $\lim\dfrac{f}{g} = \lim\dfrac{f'}{g'}$ if the second limit exists.
• Check the form first. Repeat only while the form stays indeterminate.
• The rule is not in the Indian school syllabus. JEE and IB Higher Level use it, and so do some A-Level Further Maths courses.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\lim_{x \to 0}\dfrac{\ln(1 + x) - x}{x^2}$ by a series.

2. Find $\displaystyle\lim_{x \to 0}\dfrac{e^x - e^{-x} - 2x}{x^3}$.

3. Use L'Hôpital's rule to find $\displaystyle\lim_{x \to 0}\dfrac{\sin x - x\cos x}{x^3}$.

4. Find $\displaystyle\lim_{x \to \infty}\dfrac{\ln x}{x}$.

5. A student applies L'Hôpital's rule to $\displaystyle\lim_{x \to 1}\dfrac{x + 2}{x^2 + 1}$ and gets $\dfrac{1}{2x}$ at $x = 1$, which is $\dfrac{1}{2}$. What is wrong?

References

• OpenStax, Calculus Volume 1, Chapter 4, section 4.8 on L'Hôpital's rule.
• OpenStax, Calculus Volume 2, Chapter 6, section 6.3 on Taylor and Maclaurin series.
• NCERT, Mathematics Textbook for Class XI, Chapter 12 for the standard limits. The rule and the series are not part of that text.