Question
$\lim _{x \rightarrow 0}(1+\sin x)^{1 / x}$ is equal to(a) 0(b) infinity(c) $e$(d) Does not exist.
Step 1
Step 1: We can rewrite the given limit as follows: \[\lim _{x \rightarrow 0}(1+\sin x)^{1 / x} = e^{\lim_{x \rightarrow 0} \frac{\ln(1+\sin x)}{x}}\] This is because $a^b = e^{b\ln a}$. Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 64 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x}$ is (a) 1 (b) $-1$ (c) e (d) Does not exist
$\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$ is (a) 1 (b) $-1$ (c) 0 (d) e
$\lim _{x \rightarrow 0}(\cos x+\sin x)^{\bar{x}}$ is equal to (A) $e$ (B) $e^{2}$ (C) $e^{-1}$ (D) 1
Watch the video solution with this free unlock.
EMAIL
PASSWORD