Question

prove that lim tend to 0. log(1+x^3)/sin^3x=1
plz answer

Answer 1

hope it will help you

Answer 2

$\lim_{x \to 0} \dfrac{\log (1+x^{3})}{\sin^3 x}=1$, proved.

Step-by-step explanation:

To prove that, $\lim_{x \to 0} \dfrac{\log (1+x^{3})}{\sin^3 x}=1$

L.H.S. $=\lim_{x \to 0} \dfrac{\log (1+x^{3})}{\sin^3 x}$

$=\lim_{x \to 0} \dfrac{\log (1+x^{3})}{\dfrac{x^3\sin^3 x}{x^3}}$

Multiplying numerator and denominator by $x^3$, we get

$=\lim_{x \to 0} \dfrac{\log (1+x^{3})}{x^{3}(\dfrac{\sin x}{x})^{3}}$

$=\lim_{x \to 0} \dfrac{\log (1+x^{3})}{x^{3}(1)^{3}}$

[ ∵ $\lim_{x \to 0} \dfrac{\sin x}{x} =1$]

$=\lim_{x \to 0} \dfrac{\log (1+x^{3})}{x^{3}}$

= 1

[ ∵ $\lim_{x \to 0} \dfrac{\log (1+x)}{x}=1$]

= 1 =R.H.S., proved.

Hence, $\lim_{x \to 0} \dfrac{\log (1+x^{3})}{\sin^3 x}=1$, proved.