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If F (X) = (X + 1)Cot X Be Continuous at X = 0, Then F (0) is Equal to (A) 0 (B) 1/E (C) E (D) None of These

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प्रश्न

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

विकल्प

  • 0

  • 1/e

  • e

  • none of these

MCQ
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उत्तर १

Suppose 

\[f\left( x \right)\]  is continuous at  \[x = 0 .\]
Given:
\[f\left( x \right) = \left( x + 1 \right)^{\text{ cot } x}\]

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

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

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

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

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

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

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

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

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

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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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Continuity - Exercise 9.4 [पृष्ठ ४३]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 8 Continuity
Exercise 9.4 | Q 7 | पृष्ठ ४३

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