Limits Exercise List 05

A list of more challenging limits exercises. Mastery of the previous lists is recommended before solving this one.

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Exercise 1

Calculate:

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

View solution

Indeterminate form $\dfrac{0}{0}$. Applying L'Hôpital's Rule three times (or using Taylor expansions):

$$\sin x = x - \frac{x^3}{6} + o(x^3), \quad \cos x = 1 - \frac{x^2}{2} + o(x^2)$$$$x\cos x = x - \frac{x^3}{2} + o(x^3)$$$$\sin x - x\cos x = \left(x - \frac{x^3}{6}\right) - \left(x - \frac{x^3}{2}\right) + o(x^3) = \frac{x^3}{3} + o(x^3)$$

Therefore:

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

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Exercise 2

Calculate:

$$\lim_{x \to 0} \left( \frac{1}{x^2} - \frac{1}{\sin^2 x} \right)$$

View solution

$$\frac{1}{x^2} - \frac{1}{\sin^2 x} = \frac{\sin^2 x - x^2}{x^2\sin^2 x}$$

Numerator: $\sin^2 x - x^2 = (\sin x - x)(\sin x + x)$. Using expansions:

$$\sin x = x - \frac{x^3}{6} + o(x^3) \implies \sin x - x = -\frac{x^3}{6} + o(x^3)$$$$\sin^2 x - x^2 = -\frac{x^4}{3} + o(x^4)$$

Denominator: $x^2\sin^2 x \sim x^4$.

Therefore:

$$\lim_{x \to 0} \frac{\sin^2 x - x^2}{x^2\sin^2 x} = \lim_{x \to 0} \frac{-\frac{x^4}{3}}{x^4} = -\frac{1}{3}$$

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Exercise 3

Calculate:

$$\lim_{x \to +\infty} \left( \sqrt{x^2 + x} - x \right)$$

View solution

Indeterminate form $\infty - \infty$. We rationalize:

$$\sqrt{x^2 + x} - x = \frac{(x^2 + x) - x^2}{\sqrt{x^2 + x} + x} = \frac{x}{\sqrt{x^2 + x} + x} = \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}$$

Therefore:

$$\lim_{x \to +\infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{1+1} = \frac{1}{2}$$

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Exercise 4

Calculate:

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

View solution

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

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

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Exercise 5

Calculate:

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

View solution

Using the Taylor expansion of $\arcsin x$:

$$\arcsin x = x + \frac{1}{6}x^3 + o(x^3)$$

Therefore:

$$x - \arcsin x = -\frac{1}{6}x^3 + o(x^3)$$$$\lim_{x \to 0} \frac{x - \arcsin x}{x^3} = -\frac{1}{6}$$

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Exercise 6

Calculate:

$$\lim_{x \to +\infty} x^2 \left( \sqrt[3]{x+1} - \sqrt[3]{x} \right)$$

View solution

We use the identity $a - b = \dfrac{a^3 - b^3}{a^2 + ab + b^2}$ with $a = \sqrt[3]{x+1}$ and $b = \sqrt[3]{x}$:

$$\sqrt[3]{x+1} - \sqrt[3]{x} = \frac{(x+1) - x}{\sqrt[3]{(x+1)^2} + \sqrt[3]{x(x+1)} + \sqrt[3]{x^2}} = \frac{1}{\sqrt[3]{(x+1)^2} + \sqrt[3]{x(x+1)} + \sqrt[3]{x^2}}$$

Multiplying by $x^2$:

$$x^2 \left( \sqrt[3]{x+1} - \sqrt[3]{x} \right) = \frac{x^2}{x^{2/3} \left( \sqrt[3]{(1+1/x)^2} + \sqrt[3]{1+1/x} + 1 \right)} = \frac{x^{4/3}}{\sqrt[3]{(1+1/x)^2} + \sqrt[3]{1+1/x} + 1}$$

As $x \to +\infty$, the denominator tends to $1+1+1=3$ and the numerator tends to $+\infty$. Therefore, the limit is $+\infty$.

Correcting the exponent: $x^2 / x^{2/3} = x^{4/3} \to +\infty$.

Answer: $+\infty$

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Exercise 7

Calculate:

$$\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{1/x^2}$$

View solution

Let $y = \left( \dfrac{\sin x}{x} \right)^{1/x^2}$. Then:

$$\ln y = \frac{1}{x^2} \ln\left( \frac{\sin x}{x} \right) = \frac{\ln(\sin x) - \ln x}{x^2}$$

Using $\dfrac{\sin x}{x} = 1 - \dfrac{x^2}{6} + o(x^2)$, we have:

$$\ln\left( \frac{\sin x}{x} \right) = \ln\left(1 - \frac{x^2}{6} + o(x^2)\right) \sim -\frac{x^2}{6}$$

Therefore:

$$\ln y \sim \frac{-x^2/6}{x^2} = -\frac{1}{6} \implies \lim y = e^{-1/6}$$

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Exercise 8

Calculate:

$$\lim_{x \to 0} \frac{\tan x - x}{x - \sin x}$$

View solution

Using expansions:

$$\tan x = x + \frac{x^3}{3} + o(x^3), \quad \sin x = x - \frac{x^3}{6} + o(x^3)$$$$\tan x - x = \frac{x^3}{3} + o(x^3), \quad x - \sin x = \frac{x^3}{6} + o(x^3)$$

Therefore:

$$\lim_{x \to 0} \frac{\tan x - x}{x - \sin x} = \frac{1/3}{1/6} = 2$$

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Exercise 9

Calculate:

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

View solution

As $x \to 1$, $x^x \to 1$. Indeterminate form $\dfrac{0}{0}$.

We write $x^x = e^{x\ln x}$. Differentiating the numerator (L'Hôpital's Rule):

$$\lim_{x \to 1} \frac{x^x - x}{x - 1} = \lim_{x \to 1} \left( x^x (\ln x + 1) - 1 \right) = 1\cdot(0+1) - 1 = 0$$

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Exercise 10

Calculate:

$$\lim_{x \to 0^+} \left( \frac{1}{x} \right)^{\sin x}$$

View solution

Let $y = \left( \dfrac{1}{x} \right)^{\sin x}$. Then:

$$\ln y = \sin x \cdot \ln\left(\frac{1}{x}\right) = \sin x \cdot (-\ln x) = -\sin x \ln x$$

We know that $\displaystyle\lim_{x \to 0^+} x\ln x = 0$, and since $\sin x \sim x$:

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