Limits of Trigonometric Functions

Limits of Trigonometric Functions sine,cosine,tan csc,sec and cot have important properties.
Let 'c' be a real number in the domain of the given trigonometric functions.

1) $\lim_{x->c}sin(x)$ = sin(c)

2) $\lim_{x->c}cos(x)$ = cos(c)

3) $\lim_{x->c}tan(x)$ = tan(c)

4) $\lim_{x->c}csc(x)$ = csc(c)

5) $\lim_{x->c}sec(x)$ = sec(c)

6) $\lim_{x->c}cot(x)$ = cot(c)

7) $\lim_{x->0}tan(x)$ = 0

8) $\lim_{x->0}sin(x)$ = 0

9) $\lim_{x->0}\frac{sinx}{x}$ = 1

10) $\lim_{x->0}\frac{1 - cosx}{x}$ = 0

11) $\lim_{x->0}\frac{tanx}{x}$ = 1

Examples on Limits of Trigonometric Functions

1) Evaluate : $\lim_{x->\pi /2}sin(x)$
Solution : According to the limits of the trigonometric functions, we can direct substitute x = $\pi /2$.
$\lim_{x->\pi /2}sin(x)$ = sin($\pi /2$)
According to the unit circle sin($\pi /2$) = 1
So, $\lim_{x->\pi /2}sin(x)$ = 1

2) Evaluate : $\lim_{x->\pi}tan(x)$
Solution : According to the limits of the trigonometric functions, we can direct substitute x = $\pi$.
$\lim_{x->\pi }tan(x)$ = tan($\pi$)
According to the unit circle tan($\pi $) = 0
So, $\lim_{x->\pi}tan(x)$ = 0

3) Evaluate : $\lim_{x->\pi}cos(3x)$
Solution : According to the limits of the trigonometric functions, we can direct substitute x = $\pi$.
$\lim_{x->\pi }cos(3x)$ = cos(3$\pi$)
According to the unit circle cos(3$\pi $) = -1
So, $\lim_{x->\pi}cos(3x)$ = -1

4) Evaluate : $\lim_{x->0}\frac{sinx}{5x}$
Solution : As we know that $\lim_{x->0}\frac{sinx}{x}$ = 1
So we will rearrange the given equation as ,

$\lim_{x->0}\frac{1}{5}\frac{sinx}{5x}$ = $\frac{1}{5} \lim_{x->0}\frac{sinx}{x}$

$\lim_{x->0}\frac{1}{5}\frac{sinx}{5x}$ = $\frac{1}{5} \times 1 $

$\lim_{x->0}\frac{sinx}{5x}$ = $\frac{1}{5}$

5) Evaluate : $\lim_{x->0}\frac{3(1-cosx)}{x}$
Solution : As we know that $\lim_{x->0}\frac{1- cosx}{x}$ = 0
So we will rearrange the given equation as ,

$\lim_{x->0}3\frac{1- cosx}{x}$ = 3 $\lim_{x->0}\frac{1- cosx}{x}$

$\lim_{x->0}3\left ( \frac{1 - cos x}{x} \right )$ = 3 $\times 0$

$\lim_{x->0}3\left ( \frac{1 - cos x}{x} \right )$= 0

6) Evaluate : $\lim_{t->0}\frac{sin3t}{2t}$
Solution : As we know that $\lim_{x->0}\frac{sinx}{x}$ = 1
So we will rearrange the given equation as ,

For that first we will multiply top and bottom by 3 and then interchange the position of 2 and 3 in the denominator
$\lim_{t->0}\frac{sin3t}{2t}$ = $\lim_{t->0}\frac{3 .sin3t}{3.2t}$

= $\lim_{t->0}\frac{3 .sin3t}{2.3t}$

= $ \lim_{t->0}\frac{3}{2}.\frac{sin3t}{3t}$

$ \lim_{t->0}\frac{3}{2}.\frac{sin3t}{3t}$ = $\frac{3}{2}.\lim_{t->0}\frac{sin3t}{3t}$

$\frac{3}{2}.\lim_{t->0}\frac{sin3t}{3t}$ = $\frac{3}{2}$.1

$\lim_{t->0}\frac{sin3t}{2t}$ = $\frac{3}{2}$

7) Evaluate : $\lim_{x->0}\frac{sin2x}{sin3x}$

Solution : Apply the quotient property of limit.
$\lim_{x->0}\frac{sin2x}{sin3x}$ = $\frac{\lim_{x->0}{sin2x}}{\lim_{x->0}{sin3x}}$

Multiply by numerator by 2x/2x, and denominator by 3x/3x.

$\lim_{x->0}\frac{sin2x}{sin3x}$ = $\frac{\lim_{x->0}{sin2x.\frac{2x}{2x}}}{\lim_{x->0}{sin3x.\frac{3x}{3x}}}$

$\frac{\lim_{x->0}{sin2x.\frac{2x}{2x}}}{\lim_{x->0}{sin3x.\frac{3x}{3x}}}$ = $\frac{\lim_{x->0}\frac{sin2x}{2x}.(2x)}{\lim_{x->0}{\frac{sin3x}{3x}.(3x)}}$

$\frac{\lim_{x->0}\frac{sin2x}{2x}.(2x)}{\lim_{x->0}{\frac{sin3x}{3x}.(3x)}}$ = $\frac{1.(2x)}{1.(3x)}$

x will get cancelled, so we get,

$\lim_{x->0}\frac{sin2x}{sin3x}$ = $\frac{2}{3}$


12th grade math

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