Question

If the function\[f\left( x \right) = \dfrac{{\tan \left( {\tan x} \right) - \sin \left( {\sin x}
\right)}}{{\tan x - \sin x}}\],\[\left( {x \ne 0} \right)\], is continuous at\[x = 0\],then the value of
\[f\left( 0 \right)\]is

Answer 1

Hint:A function\[f\left( x \right)\] is continuous at\[x = 0\]if \[\mathop {\lim }\limits_{x \to
0} f\left( x \right) = f\left( 0 \right)\]

In this question since the given function is continuous then we will find the value of the function at
\[\mathop {\lim }\limits_{x \to 0} \]and then we will equate the function with zero to find the value
of\[f\left( x \right)\]at which it will be continuous.

Complete step by step solution:
Given the function \[f\left( x \right) = \dfrac{{\tan \left( {\tan x} \right) - \sin \left( {\sin x}
\right)}}{{\tan x - \sin x}}\]

Given the function is continuous at \[x = 0\]and as we know when a function is continuous at \[x =
0\]then\[\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\], hence we can write the function
\[f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan \left( {\tan x} \right) - \sin \left( {\sin
x} \right)}}{{\tan x - \sin x}}\]

Now if we substitute the value of \[x = 0\]in the function, we will get the function in the form\[\left(
{\dfrac{0}{0}} \right)\]so we multiply the numerator of the function by\[\dfrac{x}{x}\], hence

we can further write this function as
\[
f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{x}{x}\left( {\tan \left( {\tan x} \right)
- \sin \left( {\sin x} \right)} \right)}}{{\tan x - \sin x}} \\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\tan \left( {\dfrac{{\tan x}}{x} \times x} \right) - \sin
\left( {\dfrac{{\sin x}}{x} \times x} \right)} \right)}}{{\tan x - \sin x}} \\
\]

Now when the limit\[x \to 0\], the value of \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} =
1\]and\[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\], hence by substituting the values we can further write the function as
\[f\left( x \right) = \dfrac{{\left( {\tan \left( {1 \times x} \right) - \sin \left( {1 \times x} \right)}
\right)}}{{\tan x - \sin x}}\]

This can also be written as

\[f\left( x \right) = \dfrac{{\left( {\tan x - \sin x} \right)}}{{\tan x - \sin x}}\]

Then by solving the function we get
\[f\left( x \right) = 1\]

Hence we can say the function will be continuous at \[x = 0\]when\[f\left( x \right) = 1\]

Therefore \[f\left( x \right) = 1\]is the answer.

Note:Limit tells the value to which a function approaches when the input of that function gets closer to some number or specific numbers.

If in a function when the numerator and the denominator both are zero when \[x = a\] then the function can be factored and simplified by cancelling. But if at \[x = a\], the denominator is equal to zero and the numerator is not zero then the limit of the function does not exist.