Function f(x) = (x + 1)^cotx will be continuous at x = 0 if the value of f(0) is

Concept:

A function is said to be continuous if 

\(\rm \displaystyle \lim_{x\rightarrow a} f(x) = f(a)\)

Calculation:

Given:

f(x) = (x + 1)cotx 

Taking log on both sides

log (f(x)) = cot x log (x + 1)

For checking continuity at x = 0

\(\rm \displaystyle \lim_{x\rightarrow 0} log(f(x)) = \rm \displaystyle \lim_{x\rightarrow 0} \cot x \ log(x\ +\ 1)\)

\(\rm \displaystyle \lim_{x\rightarrow 0} log(f(x)) = \rm \displaystyle \lim_{x\rightarrow 0} \frac { \ log(x\ +\ 1)}{tan \ x}\) = 0/0 form

Using L-Hospital rule

\(\rm \displaystyle \lim_{x\rightarrow 0} log(f(x)) = \rm \displaystyle \lim_{x\rightarrow 0} \frac { 1}{sec^2 \ x\ (x+1)} \)

\(= \rm \displaystyle \lim_{x\rightarrow 0} \frac {cos^2x }{(x+1)} \)

⇒ log (f(0)) = 1

⇒ f(0) = e¹ = e