First slide

Introduction to limits

Question

$lim_{x \to \infty} {\{{(\frac{x}{x + 1})}^{a} + \sin \frac{1}{x}\}}^{x}$ is equal to

Moderate

A

$e^{2 - 1}$
B

$e^{1 - n}$
C

$e$
D

0

Solution

We have,
$\begin{array}{l} lim_{x \to \infty} {\{{(\frac{x}{x + 1})}^{a} + \sin \frac{1}{x}\}}^{x} \\ = lim_{x \to \infty} {\{\frac{1}{{(1 + \frac{1}{x})}^{a}} + \sin \frac{1}{x}\}}^{x} \\ = lim_{x \to \infty} {\{1 + {(1 + \frac{1}{x})}^{- a} + \sin \frac{1}{x} - 1\}}^{x} \end{array}$
$\begin{array}{l} = lim_{y \to 0} {\{1 + (1 + y)^{- a} + \sin y - 1\}}^{1 / y}, where y = \frac{1}{x} \\ = lim_{y \to 0} \frac{(1 + y)^{- a} + \sin y - 1}{y} \\ = e^{y \to 0} \{\frac{\sin y}{y} + \frac{(1 + y)^{- a} - 1}{y}\} \\ = e^{1 - a} [∵ lim_{y \to 0} \frac{\sin y}{y} = 1 and lim_{y \to 0} \frac{(1 + y)^{- a} - 1}{y} = - a] \end{array}$

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