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I want to calculate this (without L'Hospital) :

$$\lim_{x \to 0} \frac{\sqrt{1+ \tan x} - \sqrt{1+ \sin x}}{x^3}$$

I already solved it in a looooong way: \begin{align} \frac{\sqrt{1+ \tan x} - \sqrt{1+ \sin x}}{x^3} &= \frac{\tan x - \sin x}{x^3} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &= \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &= \frac{\sin x (1- \cos x )}{x^3 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &= \frac{\sin x}{x} \frac{1- \cos x }{x^2 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &= \frac{\sin x}{x} \frac{(1- \cos x)(1+\cos x) }{x^2 \cos x \ (1+\cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &= \frac{\sin x}{x} \frac{1- \cos^2 x }{x^2} \frac{1}{\cos x \ (1 + \cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &= \frac{\sin x}{x} \frac{\sin^2 x }{x^2} \frac{1}{\cos x \ (1 + \cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}} \\ \\ &= 1 \cdot 1^2 \cdot \frac{1}{1 \cdot (1 +1)} \frac{1}{\sqrt{1+0}+\sqrt{1+0}} \tag{as $x \to 0$} \\ \\ &= \frac 1 4 \end{align}

Is there a shorter way?

EDIT: I am wondering whether there is a trick like $x= \arctan u^4$ etc.

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  • $\begingroup$ Do you consider asymptotic analysis tantamount to applying L'Hospital's Rule? $\endgroup$
    – Mark Viola
    Commented Oct 8, 2015 at 18:34
  • $\begingroup$ @Dr.MV Let's keep this at math 101 level $\endgroup$ Commented Oct 8, 2015 at 18:44
  • $\begingroup$ Your approach is correct and is perhaps the shortest route to answer, but it visually looks loooooong enough. you should add $\lim_{x \to 0}$ in each statement and further replace the $1/(\sqrt{1 + \tan x} + \sqrt{1 + \sin x})$ with $1/2$ in the second line itself. While evaluating limits one should replace sub-expressions with their limits whenever this is justified. $\endgroup$
    – Paramanand Singh
    Commented Oct 9, 2015 at 4:57

1 Answer 1

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Your proof looks correct (but you slightly abused notation at the end - you should say $\to 1 \cdot 1^2 \cdots$, not $= 1 \cdot 1^2 \cdots$). You can shorten it if you use $\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$.

$$\frac{\sin x}{x} \frac{1- \cos x }{x^2 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}$$

$$=\frac{\sin x}{x}\frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x\left(\sqrt{1+\tan x}+\sqrt{1+\sin x}\right)}$$

$$\stackrel{x\to 0}\to 1\cdot \frac{1}{2}\cdot \frac{1}{1\cdot (\sqrt{1+0}+\sqrt{1+0})}=\frac{1}{4}$$

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