# Evaluate $\displaystyle \Large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\sqrt{1-\tan x}}{\sin x}$

$x$ is a literal number and represents an angle of the right angled triangle as well. It formed a function in ratio form with trigonometric functions sine and tangent. The value of the function is required to evaluate when the value of $x$ tends to $0$.

$\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1-\sqrt{1-\tan x}}{\sin x}$

01

### Substitute x is equal to 0

$= \,\,\,$ $\dfrac{1-\sqrt{1-\tan (0)}}{\sin (0)}$

$= \,\,\,$ $\dfrac{1-\sqrt{1-0}}{0}$

$= \,\,\,$ $\dfrac{1-\sqrt{1}}{0}$

$= \,\,\,$ $\dfrac{1-1}{0}$

$= \,\,\,$ $\dfrac{0}{0}$

The function become indeterminate when the value of $x$ approaches zero. So, the limit problem must be solved in another method.

02

### Apply Rationalizing method

Use rationalising method by multiplying both numerator and denominator of the function by the rationalising factor of the numerator.

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1-\sqrt{1-\tan x}}{\sin x}$ $\times$ $\dfrac{1+\sqrt{1-\tan x}}{1+\sqrt{1-\tan x}}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{ (1-\sqrt{1-\tan x}) \times (1+\sqrt{1-\tan x}) }{ \sin x (1+\sqrt{1-\tan x})}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{ {(1)}^2 -{(\sqrt{1-\tan x})}^2}{\sin x (1+\sqrt{1-\tan x})}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1 -(1-\tan x)}{\sin x (1+\sqrt{1-\tan x})}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1-1+\tan x}{\sin x (1+\sqrt{1-\tan x})}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\require{cancel} \dfrac{\cancel{1}-\cancel{1}+\tan x}{\sin x (1+\sqrt{1-\tan x})}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{\tan x}{\sin x (1+\sqrt{1-\tan x})}$

03

### Use Reciprocal identity of sine and cosine

There is a $\sin x$ term in denominator and it can be cancelled by expressing $\tan x$ in terms of ratio of $\sin x$ to $\cos x$ by the reciprocal identity of sine and cosine functions.

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{\dfrac{\sin x}{\cos x}}{\sin x (1+\sqrt{1-\tan x})}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{\sin x}{\sin x \times \cos x \times (1+\sqrt{1-\tan x})}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\require{cancel} \dfrac{\cancel{\sin x}}{\cancel{\sin x} \times \cos x \times (1+\sqrt{1-\tan x})}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0}$ $\dfrac{1}{\cos x (1+\sqrt{1-\tan x})}$

04

### Substitute x is equal to 0

Now substitute $x$ is equal to zero to evaluate the function when $x$ tends to zero.

$=\,\,\,$ $\dfrac{1}{\cos (0) (1+\sqrt{1-\tan (0)})}$

$=\,\,\,$ $\dfrac{1}{1 \times (1+\sqrt{1-0})}$

$=\,\,\,$ $\dfrac{1}{1 \times (1+\sqrt{1})}$

$=\,\,\,$ $\dfrac{1}{1 \times (1+1)}$

$=\,\,\,$ $\dfrac{1}{1 \times 2}$

$=\,\,\,$ $\dfrac{1}{2}$

It is the required solution for this limit problems of the calculus mathematics.

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