Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{e^{x}+e^{-x}-2}{x\tan x}$

Answer 1

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Answer 2

we have to evaluate, $\rm \displaystyle \lim_{x \to 0}\ \frac{e^{x}+e^{-x}-2}{x\tan x}$

This question can be solved easily with help of L-Hospital rule,

check the form of limit,

putting x = 0, f(x) = {e^x + e^-x - 2}{xtanx}

f(0) = (e^0 + e^-0 - 2)/(0tan(0))

= (1 + 1 - 2)/0 = 0/0 [ so we can apply L-HOSPITAL rule

now differentiating numerator and denominator individually,

$\displaystyle\lim_{x\to 0}</p><p>\frac{e^x-e^-x}{tanx+xsec^2x}$

again, now putting, x = 0

we get, 0/0

so, again differentiating numerator and denominator individually,

$\displaystyle\lim_{x\to 0}\frac{e^x+e^-x}{sec^2x+sec^2x+2x secx . tanx}$

now putting, x = 0

= (e^0 + e^0)/{sec^2(0) + sec^2(0) + 0. sec(0) . tan(0)}

= (1 + 1)/(1 + 1 + 0)

= 2/2

= 1 [Ans]