Evaluate this super tough limit

[autodidude]

Homework Statement

Evaluate the limit as x->0 of \frac{{\sqrt{1+tan(x)}-\sqrt{1+sin(x)}}{x^3} (from Stewart's Calculus book)

The Attempt at a Solution

I've tried multiplying top and bottom by the conjugate of the numerator, tried some trig substitutions but haven't got anywhere. Even tried l'hospital's rule even though Stewart didn't intend for the reader to use it but I gave up trying to compute the 3rd derivative manually and the calculator confirms that the answer is 1/4 (I plugged in the problem as a limit and I'm not even sure it computed it before it shut off)

I suspect a sneaky substitution is required somewhere but I haven't the slightest clue what it might be as the chapter offers no clues

 

[clamtrox]

One useful thing to remember: if \lim f(x) = a and \lim g(x) = b exist and are finite, then \lim f(x) g(x) = ab. Use this after you've used the conjugate trick. Then it should be straightforward.

 

[Dickfore]

Because your limit does not render correctly, should it be:
<br /> \lim_{x \rightarrow 0}{\frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}}<br />

If yes, then I suggest you Taylor expand the numerator up to terms O(x^3). For example:
<br /> \tan x = a x + b x^3 + O(x^5)<br />
<br /> \sin x = x - \frac{x^3}{6} + O(x^5)<br />
<br /> = \tan x \, \cos x = \left( a x + b x^3 + O(x^5) \right) \left(1 - \frac{x^2}{2} + O(x^4) \right)<br />
<br /> = a x + (b - \frac{a}{2}) x^3 + O(x^5)<br />
<br /> a = 1, \ b - \frac{a}{2} = -\frac{1}{6} \Rightarrow b = \frac{1}{3}<br />


Then:
<br /> \sqrt{1 + \tan x} = \left( 1 + x + \frac{x^3}{3} + O(x^5) \right)^\frac{1}{2}<br />
<br /> = 1 + \frac{1}{2} \left( x + \frac{x^3}{3} + O(x^5) \right)<br />
<br /> + \frac{\frac{1}{2} \left(-\frac{1}{2}\right)}{2} \left(x + O(x^3)\right)^2<br />
<br /> + \frac{\frac{1}{2} \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right)}{6} \left(x + O(x^3) \right)^3 + O(x^4)<br />
<br /> = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{11 x^3}{48} + O(x^4)<br />
As a check, see here.

Do the same for \sqrt{1 + \sin x}, subtract the two and see what you get.

 

[autodidude]

@clamtrox: Thanks, I'll try that

@Dickfore: Yeah, that's it. Cheers for fixing it up. I'm not up to Taylor expansion yet, the question was in the first part of the book near the introduction to limits!

 

[Dickfore]

OK, then, after you multiply and divide by the sum of the two square root terms in the numerator, your numerator becomes:
<br /> \tan x - \sin x = \frac{\sin x \, (1 - \cos x)}{\cos x} = \frac{4 \, \sin^3\left(\frac{x}{2}\right) \, \cos\left(\frac{x}{2}\right)}{\cos x}<br />
where I used the half-angle formula.

Your 0/0 indeterminate form contains a power of \sin x/x, and you should know this limit.

 

[AGNuke]

This question doesn't require to indulge in Taylor's expansion. Its approach is quite simple in fact. Just a simple rationalization.\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}\times \frac{\sqrt{1+\tan x}+\sqrt{1+\sin x}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}
\Rightarrow\frac{(1+\tan x)-(1+\sin x)}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}
\Rightarrow \frac{\sin x(1-\cos x)}{x^3(\cos x)}\times \frac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}

You must know that
\underset{x \to 0 }{\lim}\frac{\sin x}{x}=1;\; \underset{x \to 0 }{\lim}\frac{1-\cos x}{x^2}=\frac{1}{2}; \; \underset{x \to 0 }{\lim}\cos x=1

You can see, you can club certain limits to calculate the overall limit. Try it.

 

[autodidude]

Wow, thanks a lot everyone! Will definitely 'verify' it so it sticks