Question

find the limit:
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Answer 1

Question :

Fine the value of limit

$\sf\lim_{x\to0}\dfrac{x+\sin3x}{x-\sin2x}$

Solution :

We have ,

$\rm\lim_{x\to0}\dfrac{x+\sin3x}{x-\sin2x}$

Clearly , this limit in the form of 0/0

Then ,apply L'Hospital's Rule

$\sf\lim_{x\to0}\:\dfrac{x+\sin3x}{x-\sin2x}=\lim _{x\:\to0}\dfrac{\frac{dx}{dx}+\frac{d(\sin3x)}{dx}}{\frac{dx}{dx}-\frac{d(\sin2x)}{dx}}$

$\sf=\lim_{x\:\to0}\:\dfrac{1+(\cos3x)\times3}{1-(\cos2x)\times2}$

$\sf=\lim_{x\:\to0}\dfrac{1+3\cos3x}{1-2\cos2x}$

Now put the value of limit, then

$\sf=\dfrac{1+3}{1-2}$

$\sf=\dfrac{4}{-1}$

$\sf=-4$

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About L'HOSPITAL'S Rule :

If f(a)=g(a)= 0 , then

$\sf \lim _{x \to \: a} \: \frac{f(x)}{g(x)} = \sf \lim _{x \to \: a} \frac{f'(x)}{g'(x)}$