f ( x ) f(x)

Calculus Level 4

If, f ( x ) = l o g { ( 1 + x ) 1 + x } − x x 2 , x ≠ 0 f(x)=\cfrac { log\quad \{ (1+x)^{ 1+x }\} -x }{ x^{ 2 } } \quad ,x\neq 0 is continuous at x=0. Then find f(0).If answer is 'n', Find the answer as ( n 2 ) (n^{ 2 }) .


The answer is 0.25.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Simply apply L hopital's rule and get the answer

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Deepanshu Gupta
Sep 6, 2014

f ( x ) = ( x + 1 ) log ⁡ ( x + 1 ) − x x 2 l o g ( 1 + x ) = x − x 2 2 + x 3 3 . . . . . . . . . ∞ l i m x ⟶ 0 = f ( 0 ) = 1 / 2 n = 0.5 n 2 = . 25 f(x)=\frac { (x+1)\log { (x+1)\quad -\quad x } }{ { x }^{ 2 } } \\ log(1+x)=\quad x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } .........\infty \\ lim\quad x\longrightarrow 0\quad =\quad f(0)\quad =\quad 1/2\\ n=0.5\quad \quad \\ { n }^{ 2 }=.25\\ .

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