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$$\lim_{x \to 0} \frac{x(1-\cos{x})}{\left(\tan{x}\right)^3}$$
I have got the answer to be 1/2. Is that correct, and is there any shortcut for the above problem.

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    Use $\tan x \sim x$. – 2017-02-03
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    I didn't get that. – 2017-02-03
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    In one line, using known limits: $$\frac{x(1 - \cos x)}{\tan^3 x} = \left( \frac{\tan x}{ x} \right)^{-1} \frac{1- \cos x}{x^2} \left( \frac{\tan x}{x} \right)^{-2} \to 1^{-1} \cdot \frac{1}{2} \cdot 1^{-2} = \frac{1}{2}$$ – 2017-02-03
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    Wow that's an awesome shortcut! I used complicated differentiation and things for this while u did this so simply. Thanks! – 2017-02-03

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$$\frac{x(1-\cos x)}{\tan^3x}=\frac x{\tan x}\cdot\frac{\sin^2x}{(1+\cos x)\tan^2x}=\frac x{\tan x}\cdot\frac{\cos^2x}{1+\cos x}\xrightarrow[x\to0]{}1\cdot\frac12=\frac12$$

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    Well u used rationalisation. Awesome. I could have done that! – 2017-02-03
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    @LokeshSangewar Once you see it you'll hardly forget it. Cheers. – 2017-02-03
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    Yeah! Cheers! ;-) ^_^ – 2017-02-03
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It is not complicated if you use equivalents $$\cos(x)\sim 1-\frac{x^2}2$$ $$\tan(x)\sim x$$ from which you can easily conclude.

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    And how do we get those equivalents? – 2017-02-03
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    @LokeshSangewar. I supposed you knew them. They are pretty standard and very often used in limits. – 2017-02-03
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    Nope I don't know them. I am still a class 10 student so I guess I don't know what r they. Please help me – 2017-02-03
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    @LokeshSangewar These are Taylor approximations. Look up Taylor series for these functions. – 2017-02-03
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    @ClaudeLeibovici well I haven't used these equivalents anywhere till now, so maybe if u tell me, iwill be able to solve problems a bit quicker. – 2017-02-03
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    @MrYouMath well I looked it up in wolfram only to be confused more. – 2017-02-03