Question

If f (x) = (x + 1)^cot x be continuous at x = 0, then f (0) is equal to 

पर्याय

• 0

• 1/e

• e

• none of these

Answer 1

Suppose 

\[f\left( x \right)\]  is continuous at  \[x = 0 .\]

Given:

\[f\left( x \right) = \left( x + 1 \right)^{\text{ cot } x}\]

\[\log f\left( x \right) = \left( \cot x \right) \left( \log \left( x + 1 \right) \right) \left[ \text{ Taking log on both sides }  \right]\]

\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \left( \cot x \right) \left( \log \left( x + 1 \right) \right)\]

\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{\tan x} \right)\]

\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \frac{\left( \frac{\log \left( x + 1 \right)}{x} \right)}{\left( \frac{\tan x}{x} \right)}\]

\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \frac{\lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{x} \right)}{\lim_{x \to 0} \left( \frac{\tan x}{x} \right)}\]

\[ \Rightarrow \log \left( \lim_{x \to 0} f\left( x \right) \right) = \frac{\lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{x} \right)}{\lim_{x \to 0} \left( \frac{\tan x}{x} \right)} \left[ \because f\left( x \right)\text{ is continuous at } x = 0 \right]\]

\[ \Rightarrow \log \left( \lim_{x \to 0} f\left( x \right) \right) = 1\]

\[ \Rightarrow \lim_{x \to 0} f\left( x \right) = e\]

\[ \Rightarrow f\left( 0 \right) = e \left[ \because f\left( x \right) \text{ is continuous at }  x = 0 \right]\]

Answer 2

For continuity at x = 0, we must have 

f(0) = `lim_("x"->0) "f"("x")`

`=lim_(x->0) ("x" + 1)^"cot x" = lim_(x->0) [(1 + "x")^(1/"x")]^("x cot x")`

`= lim_("x"->0)[(1 + "x")^(1/"x")]^(lim_("x"->0)("x"/("tan x"))) = "e"^1 = e`