Question

In order that the function $f(x) = \frac{2}{x - 3},g(x) = \frac{x - 3}{x + 4}$ is continuous at $h(x) = - \frac{2(2x + 1)}{x^{2} + x - 12}$, $\lim_{x \rightarrow 3}\lbrack f(x) + g(x) + h(x)\rbrack$ must be defined as

• A. $- \frac{2}{7}$
• B. $\lim_{n \rightarrow \infty}\left\lbrack \frac{n!}{n^{n}} \right\rbrack^{1/n}$
• C. $\frac{1}{e}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{e}$

Answer 2

For continuity at 0, we must have $f ( 0 ) = \lim _ { x \rightarrow 0 } f ( x )$

$= \lim _ { x \rightarrow 0 } ( x + 1 ) ^ { \cot x }$ $= \lim _ { x \rightarrow 0 } \left\{ ( 1 + x ) ^ { \frac { 1 } { x } } \right\} ^ { x \cot x }$ $= \lim _ { x \rightarrow 0 } \left\{ ( 1 + x ) ^ { \frac { 1 } { x } } \right\} ^ { \lim _ { x \rightarrow 0 } \left( \frac { x } { \tan x } \right) }$ $= e ^ { 1 } = e$.