CUET Preparation Today
CUET
-- Mathematics - Section B1
Continuity and Differentiability
$\lim\limits_{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$
0
1
$e^4$
$e^5$
$1^{\infty}$ form, given limit equal to
$e^{\lim\limits_{X \rightarrow 0}(x+4) \frac{5}{x+1}}=e^5$
Hence (4) is the correct answer.
