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Answer by Jaideep Khare for How to solve this limit question without using...

$$\lim_{x\rightarrow 0} \frac{\sin x-x}{\tan^3 x}= \lim_{x\rightarrow 0} \frac{\sin x-x}{x^3} \underbrace{\left(\frac {x^3}{\tan^3 x}\right)}_{=1}$$Now, let $$\mathrm L= \lim_{x\rightarrow 0}...

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How to solve this limit question without using L'Hopital's rule?

Find the limit $$\lim_{x\rightarrow 0} \frac{\sin x-x}{\tan^3 x}$$I found the limit which is $-\frac{1}{6}$ by using L'Hopital Rule. Is there another way to solve it without using the rule? Thanks in...

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