Practice Problems on Limits

In this page, a set of practice problems on limits are attached along with limit formulas. These solved problems on limits will be very useful to review before the final exam.

Trigonometric Limits Formulas

We will compute the limits of trigonometric functions. A few examples of trigonometric functions are sin x, cos x, tan x, sin(x+a), $\dfrac{\sin x}{x}$ etc. At first, recall the well-known formulas of trigonometric limits.

1. lim_x→0 sin x = 0

2. lim_x→0 cos x = 1

3. lim_x→0 $\dfrac{\sin x}{x}=1$

Also Read: List of all Limit Formulas

The above Formula 3 is one of the main tools to handle trigonometric limits. From this formula, we can deduce other trigonometric limits.

For example, lim_x→0 $\dfrac{\tan x}{x}$ = lim_x→0 $\dfrac{\frac{\sin x}{\cos x}}{x}$ = lim_x→0 $\dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}$ = lim_x→0 $\dfrac{\sin x}{x} \cdot$ lim_x→0 $\dfrac{1}{\cos x}$ = $1 \cdot \dfrac{1}{\cos 0}=1$.

Using Formula 1 & 2, we can obtain the limit of tan x as x tends to 0. Let’s see how we achieve this.

lim_x→0 $\tan x$ = lim_x→0 $\dfrac{\sin x}{\cos x}$ $=\dfrac{\lim\limits_{x \to 0} \sin x}{\lim\limits_{x \to 0} \cos x}$ $=\dfrac{\sin 0}{\cos 0}=\dfrac{0}{1}=0.$

Now we solve many trigonometric limits. First, we calculate the limit of $\dfrac{\sin 2x}{x}$ as x tends to zero.

Solved Problems on Trigonometric Limits

Problem 1:  Evaluate limx→0 $\dfrac{\sin 2x}{x}$

Solution:

Let z=2x. Then z→0 as x→0.

∴  lim_x→0 $\dfrac{\sin 2x}{x}$

= lim_x→0 $\left(\dfrac{\sin 2x}{2x} \cdot 2 \right)$

= 2 lim_z→0 $\dfrac{\sin z}{z}$

= 2 × 1, using the above limit formula 3: lim_x→0 $\dfrac{\sin x}{x}$=1.

= 2

Therefore, the value of the limit of sin2x/x is equal to 2 as x approaches 0.

Problem 2:  Evaluate limx→0 $\dfrac{\sin^{-1} x}{x}$

Solution:

Let z=sin^-1x.

So z=sin x. Then z → 0 as x → 0.

Now, lim_x→0 $\dfrac{\sin^{-1} x}{x}$ = lim_x→0 $\dfrac{z}{\sin z}$ $=\dfrac{1}{\lim\limits_{z \to 0}\dfrac{\sin z}{z}}$ $=\dfrac{1}{1}=1$.

So the limit of $\dfrac{\sin^{-1} x}{x}$ is equal to 1 as x tends to 0.

Problem 3:  Evaluate $\lim\limits_{x \to \frac{\pi}{2}} \dfrac{\cos x}{\frac{\pi}{2}-x}$

Solution:

Put z=$\dfrac{\pi}{2}-x$.

So x=$\dfrac{\pi}{2}-z$.

Note that z → 0 as x → $\dfrac{\pi}{2}$.

∴ $\lim\limits_{x \to \frac{\pi}{2}} \dfrac{\cos x}{\frac{\pi}{2}-x}$

= lim_z→0 $\dfrac{\cos (\dfrac{\pi}{2}-z)}{z}$

= lim_z→0 $\dfrac{\sin z}{z}$ [$\because \cos(\dfrac{\pi}{2}-x)=\sin x$]

= 1  using the formula lim_x→0 $\dfrac{\sin x}{x}=1$.

So the limit of $\dfrac{\cos x}{\frac{\pi}{2}-x}$ is equal to 1 when x tends to π/2.

Problem 4:  Evaluate limx→0 $\dfrac{\sin ax}{\sin bx}$

Solution:

lim_x→0 $\dfrac{\sin ax}{\sin bx}$

= lim_x→0 $\left(\dfrac{\sin ax}{\sin bx} \cdot \dfrac{bx}{ax} \cdot \dfrac{a}{b} \right)$

= lim_x→0 $\left(\dfrac{\sin ax}{ax} \cdot \dfrac{bx}{\sin bx} \cdot \dfrac{a}{b} \right)$

= lim_x→0 $\dfrac{\sin ax}{ax} \cdot$ lim_x→0 $\dfrac{1}{\dfrac{\sin bx}{bx}} \cdot$ lim_x→0 $\dfrac{a}{b}$

Take z=ax and w=bx. Then both z → 0 and w → 0 as x → 0.

So the above limit is equal to

lim_z→0 $\dfrac{\sin z}{z} \cdot$ lim_w→0 $\dfrac{1}{\dfrac{\sin w}{w}} \cdot$ lim_x→0 $\dfrac{a}{b}$

= lim_z→0 $\dfrac{\sin z}{z} \cdot$ $\dfrac{1}{\lim\limits_{w \to 0}\dfrac{\sin w}{w}} \cdot$ lim_x→0 $\dfrac{a}{b}$

Using lim_x→0 $\dfrac{\sin x}{x}=1,$ the above limit = $1 \cdot \dfrac{1}{1} \cdot \dfrac{a}{b}$ $=\dfrac{a}{b}$ 

Therefore, the limit of sinax/sinbx is equal to a/b as x approaches 0.

Problem 5:  Evaluate limx→0 $\dfrac{\tan x^\circ}{x}$

Solution:

Note that 180° = π.

So x°=$\dfrac{\pi x}{180}$.

Therefore, lim_x→0 $\dfrac{\tan x^\circ}{x}$

= lim_x→0 $\dfrac{\tan \dfrac{\pi x}{180}}{x}$

= lim_x→0 $\dfrac{\sin \dfrac{\pi x}{180}}{x \cos \dfrac{\pi x}{180}}$

= lim_x→0 $\dfrac{\sin \dfrac{\pi x}{180}}{x} \cdot$ lim_x→0 $\dfrac{1}{\cos \dfrac{\pi x}{180}}$

= lim_x→0 $\dfrac{\sin \dfrac{\pi x}{180}}{\dfrac{\pi x}{180}} \cdot \dfrac{\pi}{180} \cdot \lim\limits_{x \to 0} \dfrac{1}{\cos \dfrac{\pi x}{180}}$

Put $z=\dfrac{\pi x}{180}.$

Then z → 0 as x → 0.

So the above given limit is equal to

$\dfrac{\pi}{180}$ lim_z→0 $\dfrac{\sin z}{z} \cdot \dfrac{1}{\cos 0}$

= $\dfrac{\pi}{180} \cdot 1 \cdot 1$ using the formula lim_x→0 $\dfrac{\sin x}{x}=1$.

= $\dfrac{\pi}{180}$ ans.

So the limit of tanx°/x is equal to π/180 when x tends to 0.

Problem 6: limx→0 $\dfrac{\sin x (1-\cos x)}{x^3}$

Solution:

We will use the formula: 1-cos x = 2 sin²(x/2).

Now, lim_x→0 $\dfrac{\sin x (1-\cos x)}{x^3}$

= lim_x→0 $[\dfrac{\sin x}{x} \cdot \dfrac{2\sin^2 \dfrac{x}{2}}{x^2}]$

= lim_x→0 $\dfrac{\sin x}{x} \cdot \lim\limits_{x \to 0} \dfrac{\sin^2 \dfrac{x}{2}}{(\dfrac{x}{2})^2} \cdot \dfrac{1}{2}$

= lim_x→0 $\dfrac{\sin^2 \dfrac{x}{2}}{(\dfrac{x}{2})^2} \cdot \dfrac{1}{2}$ because lim_x→0 (sinx/x)=1.

= $\dfrac{1}{2} \left[\lim\limits_{x \to 0} \dfrac{\sin \frac{x}{2}}{\frac{x}{2}} \right]^2$

Put z=x/2.

So z → 0 as x → 0.

Thus, the above limit

= $\dfrac{1}{2}[\lim\limits_{z \to 0} \dfrac{\sin z}{z}]^2$

= $\dfrac{1}{2} \cdot 1^2=\dfrac{1}{2} \quad$ ans.

Problem 7:  Evaluate limx→0 $\dfrac{\cos(x+h)-\cos x}{h}$

Solution:

We will use the formula: $\cos a – \cos b=2 \sin \dfrac{a+b}{2} \cdot \sin \dfrac{b-a}{2}$

Now $L=\lim\limits_{h \to 0} \dfrac{\cos(x+h)-\cos x}{h}$ $=\lim\limits_{h \to 0} \dfrac{2 \sin \dfrac{2x+h}{2} \cdot \sin \dfrac{x-x-h}{2}}{h}$

$=\lim\limits_{h \to 0} \dfrac{2 \sin (x+\dfrac{h}{2}) \cdot \sin \dfrac{-h}{2}}{h}$ $=\lim\limits_{h \to 0} \dfrac{\sin (x+\dfrac{h}{2}) \cdot \sin \dfrac{-h}{2}}{\dfrac{h}{2}}$

$=-\lim\limits_{h \to 0} \sin (x+\dfrac{h}{2}) \cdot \dfrac{\sin \dfrac{h}{2}}{\dfrac{h}{2}} \quad$ $[\because \sin(-x)=-\sin x]$

$=-\lim\limits_{h \to 0} \sin (x+\dfrac{h}{2}) \cdot \lim\limits_{h \to 0} \dfrac{\sin \dfrac{h}{2}}{\dfrac{h}{2}}$

Put $z=\dfrac{h}{2}$. Then $z \to 0$ as $x \to 0$.

$\therefore L=-\sin(x+0) \cdot \lim\limits_{z \to 0} \dfrac{\sin z}{z}$ $=-\sin x \cdot 1$ $=-\sin x \quad$ ans.

[Formula used: $\lim\limits_{x \to 0}\dfrac{\sin x}{x}=1$]