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I am trying to find an easy way to compute the limit as $x \to 0$ of

$$f(x) = \frac{\sqrt{1+\tan(x)} - \sqrt{1+\sin(x)}}{x^3}$$

from first principles (i.e. without using l'Hôspital's rule).

I have gone as far as boiling down the problem to computing the limit as $x \to 0$ of

$$\frac{1 - \cos(x)}{x^2}$$

I thought about using the Small Angle Approximation for cosine, which indeed gives the right answer but doesn't seem to be a very formal.

Any hint?

Also, my working was fairly long so if you have a straightforward way to compute the limit of $f(x)$ I would love to hear it :)

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  • $\begingroup$ It is a bit unclear whether this post is mainly about limit of $f(x) = \frac{\sqrt{1+\tan(x)} - \sqrt{1+\sin(x)}}{x^3}$ or mailnly about limit of $\frac{1-\cos x}{x^2}$. If it is the latter, you can have a look here or here. (And there are probably several other posts about this limit, these two are the ones I was able to find quickly.) $\endgroup$ – Martin Sleziak Sep 24 '16 at 6:49
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$$\frac{1-\cos(x)}{x^2}$$ $$=\frac{2\sin^2(x/2)}{x^2}$$ $$\to ?$$

For $x$ small enough , we have:

$$\tan x>x>\sin x$$(using the geometric interpretation)

then

$$\cos(x)<\frac{\sin x}{x}<1$$

and since the function $\cos x$ is a continuous function $$\lim_{x\to 0}\cos x=cos 0=1$$

apply this to above, also recall squeeze theorem, we get what you want.

How to prove $\cos x$ is continunous, you may ask. $$|\cos x-\cos y|=2|\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})|\le2|\sin(\frac{x-y}{2})|$$ now, we only have to prove $$\lim_{x\to 0}\sin x=0$$ for x small enough,

$$0<|\sin x|<|x|$$

then

$$0\le \lim_{x\to 0}|\sin x|\le \lim_{x\to 0}|x|=0$$

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  • $\begingroup$ Thanks! How would you compute this limit though? using the small angle approximation sin(x) ~= x? $\endgroup$ – Aegis Apr 8 '14 at 14:40
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    $\begingroup$ We have $\frac{1-\cos x}{x^2}=\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)}=\frac{1}{1+\cos x}\frac{\sin^2 x}{x^2}$. $\endgroup$ – André Nicolas Apr 8 '14 at 15:09
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Try using double-angle formulae.

$1 - \cos x = 1 - \left ( 1- 2 \sin^2 \left ( \frac{\theta}{2} \right ) \right ) = 2 \sin^2 \left ( \frac{\theta}{2} \right )$

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