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.
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