Limits Exercise List 04

In this list, we work with limits involving exponentials, logarithms, and L'Hôpital's Rule.

Important reminders:

• $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
• $\displaystyle\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$
• $\displaystyle\lim_{x \to +\infty} \left(1 + \frac{1}{x}\right)^x = e$
• L'Hôpital's Rule: if $\displaystyle\lim \frac{f(x)}{g(x)}$ has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\displaystyle\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$ when the limit on the right exists.

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Exercise 1

Calculate:

$$\lim_{x \to 0} \frac{e^x - 1}{x}$$

View solution

This is a fundamental limit:

$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$

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Exercise 2

Calculate:

$$\lim_{x \to 0} \frac{e^{3x} - 1}{x}$$

View solution

$$\lim_{x \to 0} \frac{e^{3x} - 1}{x} = \lim_{x \to 0} \frac{e^{3x} - 1}{3x} \cdot 3 = 1 \cdot 3 = 3$$

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Exercise 3

Calculate:

$$\lim_{x \to 0} \frac{\ln(1 + 2x)}{x}$$

View solution

$$\lim_{x \to 0} \frac{\ln(1 + 2x)}{x} = \lim_{x \to 0} \frac{\ln(1 + 2x)}{2x} \cdot 2 = 1 \cdot 2 = 2$$

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Exercise 4

Calculate:

$$\lim_{x \to +\infty} \left(1 + \frac{3}{x}\right)^x$$

View solution

$$\lim_{x \to +\infty} \left(1 + \frac{3}{x}\right)^x = \lim_{x \to +\infty} \left(1 + \frac{3}{x}\right)^{\frac{x}{3} \cdot 3}$$$$= \left( \lim_{x \to +\infty} \left(1 + \frac{3}{x}\right)^{\frac{x}{3}} \right)^3 = e^3$$

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Exercise 5

Calculate:

$$\lim_{x \to 0} \frac{e^x - e^{-x}}{x}$$

View solution

Indeterminate form $\dfrac{0}{0}$. We apply L'Hôpital's Rule:

$$\lim_{x \to 0} \frac{e^x - e^{-x}}{x} = \lim_{x \to 0} \frac{e^x + e^{-x}}{1} = e^0 + e^0 = 1 + 1 = 2$$

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Exercise 6

Calculate:

$$\lim_{x \to 0} \frac{\ln(1 + x) - x}{x^2}$$

View solution

Indeterminate form $\dfrac{0}{0}$. We apply L'Hôpital's Rule:

$$\lim_{x \to 0} \frac{\ln(1 + x) - x}{x^2} = \lim_{x \to 0} \frac{\dfrac{1}{1+x} - 1}{2x}$$$$= \lim_{x \to 0} \frac{\dfrac{1 - (1+x)}{1+x}}{2x} = \lim_{x \to 0} \frac{-x}{2x(1+x)} = \lim_{x \to 0} \frac{-1}{2(1+x)} = -\frac{1}{2}$$

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Exercise 7

Calculate:

$$\lim_{x \to +\infty} \frac{\ln x}{x}$$

View solution

Indeterminate form $\dfrac{\infty}{\infty}$. We apply L'Hôpital's Rule:

$$\lim_{x \to +\infty} \frac{\ln x}{x} = \lim_{x \to +\infty} \frac{1/x}{1} = \lim_{x \to +\infty} \frac{1}{x} = 0$$

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Exercise 8

Calculate:

$$\lim_{x \to +\infty} \frac{x^2}{e^x}$$

View solution

Indeterminate form $\dfrac{\infty}{\infty}$. We apply L'Hôpital's Rule twice:

$$\lim_{x \to +\infty} \frac{x^2}{e^x} = \lim_{x \to +\infty} \frac{2x}{e^x} = \lim_{x \to +\infty} \frac{2}{e^x} = 0$$

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Exercise 9

Calculate:

$$\lim_{x \to 0^+} x \ln x$$

View solution

Indeterminate form $0 \cdot (-\infty)$. We rewrite it as a quotient:

$$\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x}$$

Now we have the form $\dfrac{-\infty}{+\infty}$. We apply L'Hôpital's Rule:

$$\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+}(-x) = 0$$

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Exercise 10

Calculate:

$$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$$

View solution

Indeterminate form $\dfrac{0}{0}$. We apply L'Hôpital's Rule twice:

$$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$$

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Exercise 11

Calculate:

$$\lim_{x \to +\infty} x^{1/x}$$

View solution

Let $y = x^{1/x}$. Then

$$\ln y = \frac{1}{x}\ln x = \frac{\ln x}{x}.$$

We have already seen that

$$\lim_{x \to +\infty} \frac{\ln x}{x} = 0,$$

therefore:

$$\lim_{x \to +\infty} \ln y = 0 \implies \lim_{x \to +\infty} y = e^0 = 1$$

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Exercise 12

Calculate:

$$\lim_{x \to 0} \frac{\tan x - x}{x^3}$$

View solution

Indeterminate form $\dfrac{0}{0}$. We apply L'Hôpital's Rule three times (or use Taylor expansions). The classical result is:

$$\lim_{x \to 0} \frac{\tan x - x}{x^3} = \frac{1}{3}$$