• Algebra
• Trigonometry
• Geometry
• Calculus

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 $0$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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.