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प्रश्न
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
उत्तर १
Suppose
\[\log f\left( x \right) = \left( \cot x \right) \left( \log \left( x + 1 \right) \right) \left[ \text{ Taking log on both sides } \right]\]
\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \left( \cot x \right) \left( \log \left( x + 1 \right) \right)\]
\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{\tan x} \right)\]
\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \frac{\left( \frac{\log \left( x + 1 \right)}{x} \right)}{\left( \frac{\tan x}{x} \right)}\]
\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \frac{\lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{x} \right)}{\lim_{x \to 0} \left( \frac{\tan x}{x} \right)}\]
\[ \Rightarrow \log \left( \lim_{x \to 0} f\left( x \right) \right) = \frac{\lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{x} \right)}{\lim_{x \to 0} \left( \frac{\tan x}{x} \right)} \left[ \because f\left( x \right)\text{ is continuous at } x = 0 \right]\]
\[ \Rightarrow \log \left( \lim_{x \to 0} f\left( x \right) \right) = 1\]
\[ \Rightarrow \lim_{x \to 0} f\left( x \right) = e\]
\[ \Rightarrow f\left( 0 \right) = e \left[ \because f\left( x \right) \text{ is continuous at } x = 0 \right]\]
उत्तर २
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`
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