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Limits Exercise List 03

In this list, we focus on limits involving trigonometric functions. Use fundamental limits, trigonometric identities, and substitutions when necessary.

Important fundamental limits:

lim⁡x→0sin⁡xx=1andlim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \text{and} \qquad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

Exercise 1​

Calculate:

lim⁡x→0sin⁡5xx\lim_{x \to 0} \frac{\sin 5x}{x}
View solution
lim⁡x→0sin⁡5xx=lim⁡x→0sin⁡5x5x⋅5=1⋅5=5\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} \frac{\sin 5x}{5x} \cdot 5 = 1 \cdot 5 = 5

Exercise 2​

Calculate:

lim⁡x→0sin⁡3xsin⁡7x\lim_{x \to 0} \frac{\sin 3x}{\sin 7x}
View solution
lim⁡x→0sin⁡3xsin⁡7x=lim⁡x→0sin⁡3x3x⋅7xsin⁡7x⋅37=1⋅1⋅37=37\lim_{x \to 0} \frac{\sin 3x}{\sin 7x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \frac{7x}{\sin 7x} \cdot \frac{3}{7} = 1 \cdot 1 \cdot \frac{3}{7} = \frac{3}{7}

Exercise 3​

Calculate:

lim⁡x→0tan⁡xx\lim_{x \to 0} \frac{\tan x}{x}
View solution
lim⁡x→0tan⁡xx=lim⁡x→0sin⁡xxcos⁡x=lim⁡x→0sin⁡xx⋅1cos⁡x=1⋅1=1\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x \cos x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \cdot 1 = 1

Exercise 4​

Calculate:

lim⁡x→01−cos⁡4xx2\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}
View solution

We use the fundamental limit

lim⁡θ→01−cos⁡θθ2=12\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}

Therefore:

lim⁡x→01−cos⁡4xx2=lim⁡x→01−cos⁡4x(4x)2⋅16=12⋅16=8\lim_{x \to 0} \frac{1 - \cos 4x}{x^2} = \lim_{x \to 0} \frac{1 - \cos 4x}{(4x)^2} \cdot 16 = \frac{1}{2} \cdot 16 = 8

Exercise 5​

Calculate:

lim⁡x→0sin⁡x−xx3\lim_{x \to 0} \frac{\sin x - x}{x^3}
View solution

This limit is more advanced. Using the Taylor expansion or applying L'Hôpital's rule three times, we obtain:

lim⁡x→0sin⁡x−xx3=−16\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac{1}{6}

(It can also be solved using the identity sin⁡x=3sin⁡x3−4sin⁡3x3\sin x = 3\sin\frac{x}{3} - 4\sin^3\frac{x}{3}, but the standard result is −16-\dfrac{1}{6}.)

Exercise 6​

Calculate:

lim⁡x→01−cos⁡xxsin⁡x\lim_{x \to 0} \frac{1 - \cos x}{x \sin x}
View solution
lim⁡x→01−cos⁡xxsin⁡x=lim⁡x→01−cos⁡xx2⋅xsin⁡x=12⋅1=12\lim_{x \to 0} \frac{1 - \cos x}{x \sin x} = \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \frac{x}{\sin x} = \frac{1}{2} \cdot 1 = \frac{1}{2}

Exercise 7​

Calculate:

lim⁡x→0tan⁡x−sin⁡xx3\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}
View solution
tan⁡x−sin⁡xx3=sin⁡xcos⁡x−sin⁡xx3=sin⁡x(1−cos⁡x)x3cos⁡x\frac{\tan x - \sin x}{x^3} = \frac{ \frac{\sin x}{\cos x} - \sin x }{x^3} = \frac{\sin x(1-\cos x)} {x^3\cos x}=sin⁡xx⋅1−cos⁡xx2⋅1cos⁡x=1⋅12⋅1=12= \frac{\sin x}{x} \cdot \frac{1-\cos x}{x^2} \cdot \frac{1}{\cos x} = 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{2}

Exercise 8​

Calculate:

lim⁡x→πsin⁡xx−π\lim_{x \to \pi} \frac{\sin x}{x - \pi}
View solution

We make the substitution u=x−πu = x - \pi. As x→πx \to \pi, we have u→0u \to 0.

sin⁡x=sin⁡(u+π)=−sin⁡u\sin x = \sin(u + \pi) = -\sin u

Therefore:

lim⁡x→πsin⁡xx−π=lim⁡u→0−sin⁡uu=−1\lim_{x \to \pi} \frac{\sin x}{x - \pi} = \lim_{u \to 0} \frac{-\sin u}{u} = -1

Exercise 9​

Calculate:

lim⁡x→0sin⁡23xx2\lim_{x \to 0} \frac{\sin^2 3x}{x^2}
View solution
lim⁡x→0sin⁡23xx2=lim⁡x→0(sin⁡3x3x⋅3)2=(1⋅3)2=9\lim_{x \to 0} \frac{\sin^2 3x}{x^2} = \lim_{x \to 0} \left( \frac{\sin 3x}{3x} \cdot 3 \right)^2 = (1 \cdot 3)^2 = 9

Exercise 10​

Calculate:

lim⁡x→01−cos⁡2xsin⁡23x\lim_{x \to 0} \frac{1 - \cos 2x}{\sin^2 3x}
View solution
lim⁡x→01−cos⁡2xsin⁡23x=lim⁡x→01−cos⁡2x(2x)2⋅(2x)2sin⁡23x\lim_{x \to 0} \frac{1 - \cos 2x}{\sin^2 3x} = \lim_{x \to 0} \frac{1 - \cos 2x}{(2x)^2} \cdot \frac{(2x)^2}{\sin^2 3x}=lim⁡x→01−cos⁡2x(2x)2⋅4⋅(3xsin⁡3x)2⋅19=12⋅4⋅1⋅19=29= \lim_{x \to 0} \frac{1 - \cos 2x}{(2x)^2} \cdot 4 \cdot \left( \frac{3x}{\sin 3x} \right)^2 \cdot \frac{1}{9} = \frac{1}{2} \cdot 4 \cdot 1 \cdot \frac{1}{9} = \frac{2}{9}

Exercise 11​

Calculate:

lim⁡x→0x⋅cot⁡x\lim_{x \to 0} x \cdot \cot x
View solution
lim⁡x→0x⋅cot⁡x=lim⁡x→0x⋅cos⁡xsin⁡x=lim⁡x→0xsin⁡x⋅cos⁡x=1⋅1=1\lim_{x \to 0} x \cdot \cot x = \lim_{x \to 0} x \cdot \frac{\cos x}{\sin x} = \lim_{x \to 0} \frac{x}{\sin x} \cdot \cos x = 1 \cdot 1 = 1

Exercise 12​

Calculate:

lim⁡x→0sin⁡x−tan⁡xx3\lim_{x \to 0} \frac{\sin x - \tan x}{x^3}
View solution
sin⁡x−tan⁡x=sin⁡x−sin⁡xcos⁡x=sin⁡x(1−1cos⁡x)=sin⁡x⋅cos⁡x−1cos⁡x\sin x - \tan x = \sin x - \frac{\sin x}{\cos x} = \sin x \left( 1-\frac{1}{\cos x} \right) = \sin x \cdot \frac{\cos x-1}{\cos x}

Therefore:

sin⁡x−tan⁡xx3=sin⁡xx⋅cos⁡x−1x2⋅1cos⁡x=1⋅(−12)⋅1=−12\frac{\sin x-\tan x}{x^3} = \frac{\sin x}{x} \cdot \frac{\cos x-1}{x^2} \cdot \frac{1}{\cos x} = 1 \cdot \left(-\frac{1}{2}\right) \cdot 1 = -\frac{1}{2}