AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{a^x-1}{\sin (x)}=\)
- A \(\log (a)\)
- B \(\frac{1}{2} \log (a)\)
- C 0
- D 1
Answer & Solution
Correct Answer
(A) \(\log (a)\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{a^x-1}{\sin x}=\frac{\lim _{x \rightarrow 0} \frac{a^x-1}{x}}{\lim _{x \rightarrow 0} \frac{\sin x}{x}}=\frac{\log _e a}{1}=\log _e a\) Hence, option (a) is correct.
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