# Evaluate this super tough limit

1. Aug 15, 2012

### autodidude

1. The problem statement, all variables and given/known data
Evaluate the limit as x->0 of $$\frac{{\sqrt{1+tan(x)}-\sqrt{1+sin(x)}}{x^3}$$ (from Stewart's Calculus book)

3. 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

2. Aug 15, 2012

### 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.

3. Aug 15, 2012

### Dickfore

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

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

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

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

4. Aug 15, 2012

### 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!

5. Aug 15, 2012

### Dickfore

OK, then, after you multiply and divide by the sum of the two square root terms in the numerator, your numerator becomes:
$$\tan x - \sin x = \frac{\sin x \, (1 - \cos x)}{\cos x} = \frac{4 \, \sin^3\left(\frac{x}{2}\right) \, \cos\left(\frac{x}{2}\right)}{\cos x}$$
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.

6. Aug 15, 2012

### 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.

7. Aug 15, 2012

### autodidude

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