Harder Limits: Quick Revision

A brief note on Harder Limits. Read it before a test. Try each question at the end before you open its answer.

Forms and e

• Indeterminate forms: $\dfrac{0}{0}$, $\dfrac{\infty}{\infty}$, $\infty - \infty$, $0 \cdot \infty$ and $1^{\infty}$. The shape alone does not give the limit.
• Not indeterminate: $\infty + \infty = \infty$ and $\dfrac{1}{0^+} = \infty$.
• The number e: $\displaystyle\lim_{x \to \infty}\left(1 + \dfrac{1}{x}\right)^x = e$ and $\displaystyle\lim_{x \to 0}(1 + x)^{1/x} = e$.
• Family: $\left(1 + \dfrac{k}{x}\right)^x \to e^k$ and $(1 + kx)^{1/x} \to e^k$.
• Rule for $1^{\infty}$: if $f \to 1$ and $g \to \infty$, then $f^g \to e^{\lim g(f - 1)}$.
• $\infty - \infty$: combine the fractions, or rationalise.

Squeeze, modulus and greatest integer

• Squeeze: $g \le f \le h$ and $g, h \to L$ give $f \to L$.
• Examples: $x\sin\dfrac{1}{x} \to 0$ and $\dfrac{\sin x}{x} \to 0$ as $x \to \infty$.
• Modulus: split $|x - a|$ into $x - a$ for $x > a$ and $-(x - a)$ for $x < a$, and compare the two limits.
• Greatest integer: $t - 1 < [t] \le t$, so $x\left[\dfrac{1}{x}\right] \to 1$ as $x \to 0$.
• Substitution: put $x = a + h$ to turn $x \to a$ into $h \to 0$.

Series and L'Hôpital

• Series for small $x$: $\sin x = x - \dfrac{x^3}{6}$, $\cos x = 1 - \dfrac{x^2}{2}$, $e^x = 1 + x + \dfrac{x^2}{2}$, $\ln(1 + x) = x - \dfrac{x^2}{2}$, $\tan x = x + \dfrac{x^3}{3}$.
• L'Hôpital: for $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, $\lim\dfrac{f}{g} = \lim\dfrac{f'}{g'}$ if the second limit exists. Differentiate the top and the bottom separately.
• Conditions: check the form first. Repeat only while the form stays indeterminate.
• Syllabus: the Indian school syllabus leaves the rule out. JEE and IB Higher Level use it, and so do some A-Level Further Maths courses.

A worked example

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

1. Series. $x - \sin x = \dfrac{x^3}{6} - \dots$
2. Divide by $x^3$. The quotient is $\dfrac{1}{6} - \dots$
3. Answer. The limit is $\dfrac{1}{6}$. L'Hôpital three times gives the same.

Practice with new numbers

A program to try

The program looks at four limits from this note at a point close to $0$.

Mistakes to avoid

• $1^{\infty} = 1$. The limit is a power of $e$.
• $\infty - \infty = 0$. Combine or rationalise first.
• L'Hôpital on a non-indeterminate form. Substitute first.
• L'Hôpital as the quotient rule. Differentiate the top and the bottom separately.
• Forgetting a side. Split a modulus or $[x]$.

Check yourself

1. Find $\displaystyle\lim_{x \to \infty}\left(1 + \dfrac{2}{x}\right)^{3x}$.

2. Find $\displaystyle\lim_{x \to 0} x\cos\dfrac{1}{x}$.

3. Find $\displaystyle\lim_{x \to 0}\dfrac{e^x - 1 - x}{x^2}$.

4. Find $\displaystyle\lim_{x \to \infty}\dfrac{x^3}{e^x}$.

5. Find $\displaystyle\lim_{x \to 1}\left(\dfrac{1}{x - 1} - \dfrac{3}{x^3 - 1}\right)$.

Sources

• OpenStax, Calculus Volume 1, Chapter 4, section 4.8 on L'Hôpital's rule.
• NCERT, Mathematics Textbook for Class XI, Chapter 12.