Question

explain step by step​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed{ \red{\tt \:\lim_{x\to0} \: \dfrac{tan \: x}{x} = 1}}$

$(2). \: \boxed{ \red{\tt \:\lim_{x\to0} \: \dfrac{sin \: x}{x} = 1}}$

$(3). \: \boxed{ \red{\tt \:1 - cos2x = {2sin}^{2}x }}$

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:\lim_{x\to0}\dfrac{ {x}^{2} + 1 - cosx }{x \: tanx}$

☆ If we substitute directly x = 0, we get indeterminant form

$= \tt \:\lim_{x\to0}\dfrac{ {x}^{2} + (1 - cosx)}{x \times \: \dfrac{tanx}{x} \times x}$

$= \tt \:\lim_{x\to0}\dfrac{ {x}^{2} + 2 {sin}^{2}\dfrac{x}{2} }{ {x}^{2} } \times \tt \:\lim_{x\to0}\dfrac{1}{\dfrac{tanx}{x} }$

$= \tt \:\lim_{x\to0}\dfrac{ {x}^{2} + 2 {sin}^{2}\dfrac{x}{2} }{ {x}^{2} } \times 1$

$= \tt \:\tt \:\lim_{x\to0}\dfrac{ {x}^{2} }{ {x}^{2} } + \lim_{x\to0}\dfrac{ 2 {sin}^{2}\dfrac{x}{2} }{ {x}^{2} }$

$\tt \:  = 1 \: + \: \tt \:\lim_{x\to0}\dfrac{ 2 {sin}^{2}\dfrac{x}{2} }{ \dfrac{ {x}^{2} }{4} \times 4}$

$\tt \:  = 1 \: + \: \tt \:\dfrac{1}{2} \lim_{x\to0}\dfrac{ {sin}^{2}\dfrac{x}{2} }{ \dfrac{ {x}^{2} }{4} }$

$\tt \:  = 1 + \dfrac{1}{2} \times {(1)}^{2}$

$\tt \:  = \dfrac{3}{2}$

$\tt\implies \:\: \boxed{ \red{\tt \:\lim_{x\to0} \: \tt \:\dfrac{ {x}^{2} + 1 - cosx }{x \: tanx} = \: \dfrac{3}{2} }}$