Question

For $x > 0$, calculate the value of ${\lim }_{x \to 0} \left[ {{{\left( {\sin x} \right)}^{\frac{1}{x}}} + {{\left( {\dfrac{1}{x}} \right)}^{\sin x}}} \right]$
A. 0
B. -1
C. 1
D. 2

Answer 1

Hint: First we will apply the formula ${\lim }_{x \to a} \left( {f\left( x \right) \pm g\left( x \right)} \right) = {\lim }_{x \to a} f\left( x \right) \pm {\lim }_{x \to a} g\left( x \right)$. Then by using the logarithm property and L'hospital, we will calculate the limit of the second term.

Formula Used:
${\lim }_{x \to a} \left( {f\left( x \right) \pm g\left( x \right)} \right) = {\lim }_{x \to a} f\left( x \right) \pm {\lim }_{x \to a} g\left( x \right)$
$\log {a^m} = m\log a$
$\ln \dfrac{1}{a} = - \ln a$

Complete step by step solution:
Given limit is ${\lim }_{x \to 0} \left[ {{{\left( {\sin x} \right)}^{\frac{1}{x}}} + {{\left( {\dfrac{1}{x}} \right)}^{\sin x}}} \right]$
Apply the formula ${\lim }_{x \to a} \left( {f\left( x \right) \pm g\left( x \right)} \right) = {\lim }_{x \to a} f\left( x \right) \pm {\lim }_{x \to a} g\left( x \right)$
$ = {\lim }_{x \to 0} {\left( {\sin x} \right)^{\frac{1}{x}}} + {\lim }_{x \to 0} {\left( {\dfrac{1}{x}} \right)^{\sin x}}$
Putting the limit in the first term
$ = {\lim }_{x \to 0} {\left( 0 \right)^{\frac{1}{x}}} + {\lim }_{x \to 0} {\left( {\dfrac{1}{x}} \right)^{\sin x}}$
$ = 0 + {\lim }_{x \to 0} {\left( {\dfrac{1}{x}} \right)^{\sin x}}$
Rewrite the in the form ${\lim }_{x \to a} f\left( x \right) = {e^{{\lim }_{x \to a} \ln \left( {f\left( x \right)} \right)}}$
$ = 0 + {e^{{\lim }_{x \to 0} \ln \left[ {{{\left( {\dfrac{1}{x}} \right)}^{\sin x}}} \right]}}$
Now apply $\log {a^m} = m\log a$
$ = 0 + {e^{{\lim }_{x \to 0} \sin x\ln \dfrac{1}{x}}}$
Again apply the formula in the second term $\ln \dfrac{1}{a} = - \ln a$
$ = 0 + {e^{{\lim }_{x \to 0} \left( { - \sin x\ln x} \right)}}$
Convert $\sin x$ into $\text{cosec }x$
$ = 0 + {e^{{\lim }_{x \to 0} \left( { - \dfrac{{\ln x}}{{\text{cosec }x}}} \right)}}$
If we put $x = 0$ in $\ln x$ and $\text{cosec }x$, then both will tend to infinity. So, we can apply the L' Hospital formula.
$ = 0 + {e^{{\lim }_{x \to 0} \left( { - \dfrac{{\dfrac{1}{x}}}{{ - \text{cosec }x\cot x}}} \right)}}$
$ = 0 + {e^{{\lim }_{x \to 0} \left( {\dfrac{{\tan x}}{x} \cdot \sin x} \right)}}$
Now applying the formula ${\lim }_{x \to a} \left( {f\left( x \right) \cdot g\left( x \right)} \right) = {\lim }_{x \to a} f\left( x \right) \cdot {\lim }_{x \to a} g\left( x \right)$
$ = 0 + {e^{{\lim }_{x \to 0} \dfrac{{\tan x}}{x} \cdot {\lim }_{x \to 0} \sin x}}$
Apply the formula ${\lim }_{x \to 0} \dfrac{{\tan x}}{x} = 1$
$ = 0 + {e^{1 \cdot \sin 0}}$
$ = 0 + {e^0}$ since $\sin 0 = 0$
Now apply the formula ${e^0} = 1$.
$ = 0 + 1 = 1$

Option ‘C’ is correct

Note: In the given question first you need to apply the sum of limit formula to break it into two limits. You can put the value of limit in the first term because you will get a value that is zero. But if you limit the value in the second term, then you will not get a finite value. So that you are unable to put the value of the limit in the second term.