JEE Advanced · Mathematics · 16. Limits
For \(x>0, \lim _{x \rightarrow 0}\left((\sin x)^{1 / x}+\left(\frac{1}{x}\right)^{\sin x}\right)\) is
- A
0
- B
\(-1\)
- C
1
- D
2
Answer & Solution
Correct Answer
(C)
1
Step-by-step Solution
Detailed explanation
Here, \(\lim _{x \rightarrow 0}(\sin x)^{1 / x}+\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \Rightarrow 0+\lim _{x \rightarrow 0} e^{\log \left(\frac{1}{x}\right)^{\sin x}} \quad\left\{\begin{array}{l}\lim _{x \rightarrow 0}(\sin x)^{1 / x} \rightarrow 0 \\ \text { as, (decimal) })^x \rightarrow 0\end{array}\right\}\) \(\Rightarrow \quad e^{\lim _{x \rightarrow 0} \frac{\log (1 / x)}{\operatorname{cosec} x}}\)
On applying L Hospital's Rule, we get
\[
\Rightarrow \quad e^{\lim _{x \rightarrow 0} \frac{x\left(-\frac{1}{x^2}\right)}{-\operatorname{cosec} x \cot x}}=e^{\lim _{x \rightarrow 0} \frac{\sin x}{x} \tan x}=e^0=1
\]
On applying L Hospital's Rule, we get
\[
\Rightarrow \quad e^{\lim _{x \rightarrow 0} \frac{x\left(-\frac{1}{x^2}\right)}{-\operatorname{cosec} x \cot x}}=e^{\lim _{x \rightarrow 0} \frac{\sin x}{x} \tan x}=e^0=1
\]
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