Question

limit X approaches to 0
log(1+X)/Sin x​

Answer 1

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

$\large\boxed{ \red{\tt \:  (1) \: \tt \:\lim_{x\to 0} \: log(\dfrac{1 + x}{x} ) \: = \: 1 }}$

$\large \boxed{ \green{\tt \:  (2) \:\tt \:\lim_{x\to 0} \: \dfrac{sinx}{x} \: = \: 1} }$

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

$\tt :  ⟼ \: \tt \:\lim_{x\to 0}\dfrac{ log(1 + x) }{sinx}$

$\tt :  ⟼\lim_{x\to 0} \: \bigg(\dfrac{ log(1 + x) }{x} \div \dfrac{x}{sinx} \bigg)$

$\tt :  ⟼\lim_{x\to 0} \: log(\dfrac{1 + x }{x} ) \div \: \lim_{x\to 0}\dfrac{x}{sinx}$

$\tt :  ⟼ \: 1 \: \times \: 1$

$\tt :  ⟼ \: 1$