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I only need help in the first limit and then I will do the rest alone and post the answer.I have to evaluate 20 limit in my homework but I only need the method and I swear I will do the rest on my own. Please just show me the way and don't downvote please.

$$ \lim_{x\to 0} \frac{(\sin x)^2\tan x }{x^4} $$

$$ \lim_{x\to 0} \frac{\tan 4x }{\sin 2x} $$

$$ \lim_{x\to 0} \frac{(1- \cos 2x) }{\tan^2 3x} $$

$$ \lim_{x\to 0} \frac{2\sin x-\sin 2x }{x^2} $$

Thank you for help.

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    Hint. Have a look at L'Hopital' s rule: https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule – 2017-02-18
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    We didn't study it yet but I'll try to do it. Thank you – 2017-02-18
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    Use the standard limit $$\lim_{x\to 0}\frac {\sin x} {x} =1$$ – 2017-02-18

3 Answers 3

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In general, for limits $x\rightarrow 0$, I would do something like this:

  1. Remove $\tan$ be writing it as $\frac\sin\cos$.
  2. Using Taylor expansions of $\cos$ and $\sin$ up to an appropriate order. Spoken simply: replace the trigonometric function by a polynomial that fits its behaviour near zero very well.

$$ \cos(x)\approx 1-\frac12 x^2+\frac1{24}x^4-\cdots $$ $$ \sin(x)\approx x-\frac16 x^3+\frac1{120}x^5-\cdots $$

You then can use your usual limit techniques for polynomials. As a rule of thumb, most of the time it suffices to know, that $\sin$ behaves like $x$ and $\cos$ behaves like $1$ or $1-\frac 12x^2$.

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    I don't know Taylor expansions, we didn't get to that yet. The teacher only gave us these three methods : $$ \lim_{x\to 0} \frac{\sin x}{x} = 1 $$ $$ \lim_{x\to 0} \frac{\tan x}{x} = 1 $$ $$ \lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac {1} {2} $$ And then he gave us this exercise – 2017-02-18
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    @M.Winter: Typo - $\tan=\frac{\sin}{\cos}$ – 2017-02-18
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All these basically boil down to the basic limit $$ \lim_{t\to0}\frac{\sin t}{t}=1 $$ Let me show the third one. Recall that $1-\cos2x=2\sin^2x$, so $$ \lim_{x\to0}\frac{1-\cos2x}{\tan^23x}= \lim_{x\to0}2\sin^2x\frac{\cos^23x}{\sin^23x}= \lim_{x\to0}\frac{2}{9}\cos^23x\frac{\sin^2x}{x^2}\frac{(3x)^2}{\sin^23x} $$

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    Thank you. I appreciate your help – 2017-02-18
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We start with

$$\lim_{x\to 0}{\sin{x}\over x}=\lim_{x\to 0}{\tan{x}\over x}=1$$

With this in mind we get

$$\lim_{x\to 0}{\sin^2{x}\tan{x}\over x^4}=\lim_{x\to 0}{\sin^2{x}\over x^2}{\tan{x}\over x}{1\over x}=\infty$$

$$\lim_{x\to 0}{\tan{4x}\over \sin{2x}}=\lim_{x\to 0}{\tan{4x}\over 4x}{2x\over \sin{2x}}{4x\over 2x}=2$$

The two remaining require $\sin{2x}=2\sin{x}\cos{x}$ and $1-\cos{2x}=2\sin^2{x}$ and so we have

$$\lim_{x\to 0}{1-\cos{2x}\over \tan^2{3x}}=\lim_{x\to 0}{2\sin^2{x}\over \tan^2{3x}}={2\over 9}$$

$$\lim_{x\to 0}{2\sin{x}-\sin{2x}\over x^2}=\lim_{x\to 0}{2\sin{x}(1-\cos{x})\over x^2}=\lim_{x\to 0}{4\sin{x}\sin^2{x\over 2}\over x^2}=0$$

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    This is definitely helpful Thank you so much – 2017-02-18