Trigonometric Limits: Problems and Solutions

We will compute the limits of trigonometric functions. A few examples of trigonometric functions are sin x, cos x, tan x, sin (x+a), frac{sin x}{x} etc. At first, recall the well-known formulas of trigonometric limits.
1. lim_{x to 0} sin x =0
2. lim_{x to 0} cos x =1
3. lim_{x to 0} frac{sin x}{x}=1
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 to 0} frac{tan x}{x} =lim_{xto 0} frac{frac{sin x}{cos x}}{x} =lim_{xto 0} frac{sin x}{x} cdot frac{1}{cos x} =lim_{xto 0} frac{sin x}{x} cdot lim_{xto 0} frac{1}{cos x} =1 cdot frac{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 to 0} tan x =lim_{x to 0} frac{sin x}{cos x} =frac{lim_{x to 0} sin x}{lim_{x to 0} cos x} =frac{sin 0}{cos 0}=frac{0}{1}=0.
Now we solve many trigonometric limits. First, we calculate the limit of frac{sin 2x}{x} as x tends to zero.
Problem 1:  Evaluate lim_{x to 0} frac{sin 2x}{x}
Solution:
Let z=2x. Then  z to 0 as x to 0.
therefore lim_{x to 0} frac{sin 2x}{x} =lim_{x to 0} (frac{sin 2x}{2x} cdot 2) =2 lim_{z to 0} frac{sin z}{z} =2 cdot 1=2 quad ans.
[Formula used: lim_{x to 0} frac{sin x}{x}=1]
Next, we calculate the limit of frac{sin^{-1} x}{x} as x tends to 0.
Problem 2:  Evaluate lim_{x to 0} frac{sin^{-1} x}{x}
Solution:
Let z=sin^{-1}x quad So  z=sin x. Then z to 0 as x to 0.
therefore lim_{x to 0} frac{sin^{-1} x}{x} =lim_{z to 0} frac{z}{sin z} =frac{1}{lim_{z to 0}frac{sinz}{z}} =frac{1}{1}=1 quad ans.
Problem 3:  Evaluate lim_{x to pi/2} frac{cos x}{pi/2-x}
Solution:
Put z=pi/2-x. So x=pi/2-z. Also, we have z to 0 as x to pi/2.
therefore lim_{x to pi/2} frac{cos x}{pi/2-x} =lim_{z to 0} frac{cos (pi/2-z)}{z} =lim_{z to 0} frac{sin z}{z} [because cos(pi/2-x)=sin x]
=1 quad ans.   [because lim_{x to 0} frac{sin x}{x}=1]
Next, we calculate the limit of sin ax / sin bx as x tends to 0.
Problem 4:  Evaluate lim_{x to 0} frac{sin ax}{sin bx}
Solution:
L=lim_{x to 0} frac{sin ax}{sin bx} =lim_{x to 0} (frac{sin ax}{sin bx} cdot frac{bx}{ax} cdot frac{a}{b}) =lim_{x to 0} (frac{sin ax}{ax} cdot frac{bx}{sin bx} cdot frac{a}{b})
=lim_{x to 0} frac{sin ax}{ax} cdot lim_{x to 0} frac{1}{frac{sin bx}{bx}} cdot lim_{x to 0} frac{a}{b}
Take z=ax and w=bx. Then both z to 0 and w to 0 as x to 0. Hence
L=lim_{z to 0} frac{sin z}{z} cdot lim_{w to 0} frac{1}{frac{sin w}{w}} cdot lim_{x to 0} frac{a}{b} =lim_{z to 0} frac{sin z}{z} cdot frac{1}{lim_{w to 0}frac{sin w}{w}} cdot lim_{x to 0} frac{a}{b}
Using lim_{x to 0} frac{sin x}{x}=1, we have L=1 cdot frac{1}{1} cdot frac{a}{b} =frac{a}{b} quad ans.
Next, we calculate the limit of frac{tan x^circ}{x} as x tends to 0.
Problem 5:  Evaluate lim_{x to 0} frac{tan x^circ}{x}
Solution:
Note that 180^circ =pi quad So  x^circ=frac{pi x}{180}.
L= lim_{x to 0} frac{tan x^circ}{x} =lim_{x to 0} frac{tan frac{pi x}{180}}{x} =lim_{x to 0} frac{sin frac{pi x}{180}}{x cos frac{pi x}{180}}
=lim_{x to 0} frac{sin frac{pi x}{180}}{x} cdot lim_{x to 0} frac{1}{cos frac{pi x}{180}} =lim_{x to 0} frac{sin frac{pi x}{180}}{frac{pi x}{180}} cdot frac{pi}{180} cdot lim_{x to 0} frac{1}{cos frac{pi x}{180}}
Put z=frac{pi x}{180}. Then z to 0 as x to 0. Hence
L=frac{pi}{180} lim_{z to 0} frac{sin z}{z} cdot frac{1}{cos 0} =frac{pi}{180} cdot 1 cdot 1 =frac{pi}{180} quad ans.
[Formula used: lim_{x to 0}frac{sin x}{x}=1]
Problem 6:  Evaluate lim_{h to 0} frac{cos(x+h)-cos x}{h}
Solution:
We will use the formula: cos a – cos b=2 sin frac{a+b}{2} cdot sin frac{b-a}{2}
Now L=lim_{h to 0} frac{cos(x+h)-cos x}{h} =lim_{h to 0} frac{2 sin frac{2x+h}{2} cdot sin frac{x-x-h}{2}}{h}
=lim_{h to 0} frac{2 sin (x+frac{h}{2}) cdot sin frac{-h}{2}}{h} =lim_{h to 0} frac{sin (x+frac{h}{2}) cdot sin frac{-h}{2}}{frac{h}{2}}
=-lim_{h to 0} sin (x+frac{h}{2}) cdot frac{sin frac{h}{2}}{frac{h}{2}} quad [because sin(-x)=-sin x]
=-lim_{h to 0} sin (x+frac{h}{2}) cdot lim_{h to 0} frac{sin frac{h}{2}}{frac{h}{2}}
Put z=frac{h}{2}. Then z to 0 as x to 0.
therefore L=-sin(x+0) cdot lim_{z to 0} frac{sin z}{z} =-sin x cdot 1 =-sin x quad ans.
[Formula used: lim_{x to 0}frac{sin x}{x}=1]
Problem 7:  Evaluate lim_{x to 0} frac{sin x (1-cos x)}{x^3}
Solution:
We will use the formula: 1-cos x=2sin^2 frac{x}{2}
Now L=lim_{x to 0} frac{sin x (1-cos x)}{x^3} =lim_{x to 0} [frac{sin x}{x} cdot frac{2sin^2 frac{x}{2}}{x^2}]
=lim_{x to 0} frac{sin x}{x} cdot lim_{x to 0} frac{sin^2 frac{x}{2}}{(frac{x}{2})^2} cdot frac{1}{2}
=lim_{x to 0} frac{sin^2 frac{x}{2}}{(frac{x}{2})^2} cdot frac{1}{2} quad [because lim_{x to 0} frac{sin x}{x}=1]
=frac{1}{2} [lim_{x to 0} frac{sin frac{x}{2}}{frac{x}{2}}]^2
Put z=frac{x}{2}. So z to 0 as x to 0. Therefore
=frac{1}{2}[lim_{z to 0} frac{sin z}{z}]^2 =frac{1}{2} cdot 1^2=frac{1}{2} quad ans.
Related Topics: