Question-112866

Question Number 112866 by bemath last updated on 10/Sep/20

Answered by bobhans last updated on 10/Sep/20

(7) L = \lim_{x \to 0} {(\frac{1}{x})}^{\tan x}

L = e^{\underset{x \to 0}{\lim t a n} x . \ln (\frac{1}{x})} = e^{\lim_{x \to 0} - \tan x . \ln x}

L = e^{\lim_{x \to 0} \frac{- \ln x}{\cot x}} = e^{\lim_{x \to 0} \frac{- \frac{1}{x}}{- cosec^{2} x}}

L = e^{\lim_{x \to 0} \frac{\sin^{2} x}{x}} = e^{0} = 1

Answered by Dwaipayan Shikari last updated on 10/Sep/20

5)

\lim_{x \to 0} \frac{e^{x} - e^{- x}}{s i n x} = \frac{e^{x} - 1}{x} + \frac{1 - \frac{1}{e^{x}}}{x} = 1 + \frac{e^{x} - 1}{e^{x} x} = 2

Answered by Dwaipayan Shikari last updated on 10/Sep/20

\lim_{x \to 0} \frac{e^{x^{2}} - 1}{1 - c o s x} = \lim_{x \to 0} \frac{e^{x^{2}} - 1}{2 {s i n}^{2} \frac{x}{2}} = 2 \frac{e^{x^{2}} - 1}{x^{2}} = 2

Answered by mathmax by abdo last updated on 10/Sep/20

5) let use hospital theorem

\lim_{x \to 0} \frac{e^{x} - e^{- x}}{sinx} = \lim_{x \to 0} \frac{2 sh (x)}{sinx} = \lim_{x \to 0} \frac{2 ch (x)}{cosx} = 2

Answered by mathmax by abdo last updated on 10/Sep/20

6) let f (x) = \frac{e^{x^{2}} - 1}{1 - cosx} we have e^{u} \sim 1 + u \Rightarrow e^{x^{2}} \sim 1 + x^{2} and

1 - cosx \sim \frac{x^{2}}{2} \Rightarrow f (x) \sim \frac{x^{2}}{\frac{x^{2}}{2}} = 2 \Rightarrow \lim_{x \to 0} f (x) = 0

Answered by mathmax by abdo last updated on 10/Sep/20

7) let g (x) = {(\frac{1}{x})}^{tanx} \Rightarrow g (x) = e^{tanxln (\frac{1}{x})} = e^{- \ln (x) tanx}

\Rightarrow g (x) \sim e^{- xlnx} \Rightarrow \lim_{x \to 0} g (x) = e^{0} = 1