Question

The value of limx →0 log sin 2x / logx is​

Answer 1

Basic Concept :-

L'Hospital's Rule

• It is used to solve the limits when we have an meaningless or indeterminate form 0/0 or ∞/∞ , then differentiate the numerator and differentiate the denominator independently and then take the limit if indeterminant form is removed otherwise repeat the process till indeterminant form is removed.

Formula used :-

$\blue{\boxed{\sf\:\dfrac{d}{dx}logx = \dfrac{1}{x}}}$

$\blue{\boxed{\sf\:\dfrac{d}{dx}sinx = cosx}}$

$\blue{\boxed{\sf\:\dfrac{d}{dx}x = 1}}$

$\blue{\boxed{\sf \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{tanx}{x} = 1}}$

Let's solve the problem now!!

$\rm :\longmapsto\: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{log \: sin2x}{log \: x}$

On directly Substitute the value of x = 0, we get

$\rm :\longmapsto\: \: = \: \tt \: \dfrac{log \: sin(2 \times 0)}{log \: 0}$

$\rm :\longmapsto\: \: = \: \tt \: \dfrac{log \:0}{ \infty }$

$\rm :\longmapsto\: \: = \: \tt \: \dfrac{ \infty }{ \infty }$

Its an indeterminant form,

So,

By applying L - Hospital Rule, we get

$\rm :\longmapsto\: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{\dfrac{d}{dx} \: log \: sin2x}{\dfrac{d}{dx} \: log \: x}$

$\rm \: = \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{\dfrac{1}{sin \: 2x} \dfrac{d}{dx} \: sin2x}{\dfrac{1}{x} }$

$\rm \: = \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{\dfrac{1}{sin \: 2x} \: cos \: 2x \: \dfrac{d}{dx} \: 2x}{\dfrac{1}{x} }$

$\rm \: = \: \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{cot2x \: \times \: 2}{\dfrac{1}{x} }$

$\rm \: = \: \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{2x}{tan \: 2x}$

$\rm \: = \: \: 1$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \purple{\underbrace{\pink{\boxed{ \bf\: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{log \: sin2x}{log \: x} = 1}}}}$

Additional Information :-

$\blue{\boxed{\sf \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{sinx}{x} = 1}}$

$\blue{\boxed{\sf \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{log(1 + x)}{x} = 1}}$

$\blue{\boxed{\sf \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{ {e}^{x} - 1}{x} = 1}}$

$\blue{\boxed{\sf \: \displaystyle \lim_{x \to \: 0} \tt \: \dfrac{ {a}^{x} - 1}{x} = log(a)}}$