Question

Please help me to solve this.​

Answer 1

$\rm \longrightarrow\lim \limits_{x \rightarrow0}{ \dfrac{ {e}^{x} - {e}^{ - x} - 2x }{x - sin \: x} } \\ \\ \rm \longrightarrow\lim \limits_{x \rightarrow0}{ \dfrac{ {e}^{x} + {e}^{ - x} - 2 }{1- cos\: x} }\\ \\ \rm \longrightarrow\lim \limits_{x \rightarrow0}{ \dfrac{ {e}^{x} - {e}^{ -x} - 0 }{0 + sin\: x} }\\ \\ \rm \longrightarrow\lim \limits_{x \rightarrow0}{ \dfrac{ {e}^{x} + {e}^{ - x} - 0 }{cos \: x} } \\ \\ \\\rm \longrightarrow \dfrac{ {e}^{0} + \dfrac{1}{{e}^{ 0}} }{cos \: 0} \\ \\ \\\rm \longrightarrow \dfrac{ 1 + 1 }{1} = 2$

While solving this question we have applied L' hospital rule wherever we are getting 0/0 form. In this we have to differentiate numerator and denominator seperately.

Additional-Information :-

While applying L' hospital rule we have to differentiate. To differentiate we must know basic differentiation formulae :-

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x