1. ## Trig Limit

Find the value of $\displaystyle \displaystyle\lim_{x\to{0}}\frac{\sqrt{1+tan\;x}-\sqrt{1+sin\;x}}{x^3}$

Is there an alternate way of doing this question without using l'Hosital's Rule

or this: $\displaystyle \displaystyle\lim_{x\to{0}}\frac{\sqrt{1+tan\;x}-\sqrt{1+sin\;x}}{x^3}\;\frac{\sqrt{1+tan\;x}+\sqrt {1+sin\;x}}{\sqrt{1+tan\;x}+\sqrt{1+sin\;x}}$ and then having to go through a long simplification before getting the answer....

2. $\displaystyle \displaystyle\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}\cdot\frac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}=\frac{1}{2}\lim_{x\to 0}\frac{\sin x(1-\cos x)}{x^3\cos x}=$
$\displaystyle \displaystyle=\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{ 4}$

3. Originally Posted by red_dog
$\displaystyle \displaystyle\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}\cdot\frac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}=\frac{1}{2}\lim_{x\to 0}\frac{\sin x(1-\cos x)}{x^3\cos x}=$
$\displaystyle \displaystyle=\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{ 4}$
When you take the limit of $\displaystyle \frac{2\sin^2\frac{x}{2}}{x^2}$ doesn't that equal 0 so then how do you get $\displaystyle \frac{1}{4}$?

4. Originally Posted by polymerase
When you take the limit of $\displaystyle \frac{2\sin^2\frac{x}{2}}{x^2}$ doesn't that equal 0 so then how do you get $\displaystyle \frac{1}{4}$?
$\displaystyle \lim_{x \to 0} \frac{2 sin^2 \left ( \frac{x}{2} \right )}{x^2}$

$\displaystyle = 2 \lim_{x \to 0} \left ( \frac{sin \left ( \frac{x}{2} \right )}{x} \right ) ^2$

Now change the variable slightly: $\displaystyle y = \frac{x}{2}$

Soooo....
$\displaystyle \lim_{x \to 0} \frac{2 sin^2 \left ( \frac{x}{2} \right )}{x^2} = 2 \lim_{y \to 0} \left ( \frac{sin(y)}{2y} \right )^2$

$\displaystyle = \frac{2}{4} \lim_{y \to 0} \left ( \frac{sin(y)}{y} \right )^2$

$\displaystyle = \frac{1}{2} (1)^2 = \frac{1}{2}$

With the other 1/2 out in front of the expression we get red_dog's answer of 1/4.

-Dan

5. Originally Posted by topsquark
$\displaystyle \lim_{x \to 0} \frac{2 sin^2 \left ( \frac{x}{2} \right )}{x^2}$
$\displaystyle \lim_{x \to 0} \frac{{2\sin ^2 \dfrac{x}{2}}}{{x^2 }} = \frac{1}{2}\lim_{x \to 0} \left( {\frac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2 = \frac{1}{2}.$

6. Originally Posted by Krizalid
He is a physicist leave him alone!

7. Originally Posted by Krizalid
$\displaystyle \lim_{x \to 0} \frac{{2\sin ^2 \dfrac{x}{2}}}{{x^2 }} = \frac{1}{2}\lim_{x \to 0} \left( {\frac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2 = \frac{1}{2}.$

I always miss the simple way...

-Dan

8. Alternatively, stepping in here

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

$\displaystyle \tan(x)-\sin(x)=\bigg[x+\frac{x^3}{3}+\cdots\bigg]-\bigg[x-\frac{x^3}{6}+\cdots\bigg]\sim\frac{x^3}{2}$

giving us

$\displaystyle \lim_{x\to{0}}\frac{\frac{x^3}{2}}{2x^3}=\frac{1}{ 4}$

9. Originally Posted by polymerase
When you take the limit of $\displaystyle \frac{2\sin^2\frac{x}{2}}{x^2}$ doesn't that equal 0 so then how do you get $\displaystyle \frac{1}{4}$?
It actually equals zero over zero, but by using L'Hopital's rule, you evaluate the zero limit of:

$\displaystyle \frac{-2\cos(\frac{x}{2})sin(\frac{x}{2})}{2x}$ which is again zero over zero, so one more time...

$\displaystyle f(x) = \frac{2\sin(\frac{x}{2})sin(\frac{x}{2}) - 2\cos^2(\frac{x}{2})}{4}$

$\displaystyle \lim_{x \rightarrow 0} f(x) = \frac{2}{4}$

Multiply that by the rest of the original limit and you have one fourth.