# Indeterminate limit problem

1. Jun 1, 2014

### khwarizm

1. The problem statement, all variables and given/known data
i need to find the value of limit but its indeterminate and i cant use l'hospital's rule.
i multiplied nominator and denominator with $\displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}{|x|}}$ to get rid of square root. After get i tried to make some transforms to simplify something but i couldn find anything right.

2. Relevant equations
This is the question:
$\displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1-cos^2(x)}}{|x|}}$

3. The attempt at a solution
this is the last step that i came.
$\displaystyle\lim_{x\rightarrow 0^+} {\frac{1-cos(x)+1-cos^{2}(x)}{|x|\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}}$

2. Jun 1, 2014

### pasmith

Does recognising that if $f(0) = 0$ then $$f'(0) = \lim_{x \to 0^{+}} \frac{f(x)-f(0)}{x} = \lim_{x \to 0^{+}} \frac{f(x)}{|x|}$$count as "using l'Hopital's rule"?

I personally would use $\cos^2(x/2) - \sin^2(x/2) = \cos \theta$ to express $2\sin^{2}(x/2)+1+\cos^2(x)$ as a quadratic in $\cos x$ and then complete the square before differentiating.

3. Jun 1, 2014

### lurflurf

$$\frac{\sqrt{2\sin^{2}(x/2)+1-\cos^2(x)}}{|x|}=\sqrt{\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2+\left(\frac{sin(x)}{x}\right)^2}$$
Do you know what
$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=\sin^\prime(0)$$
is?

I would say recognizing
$$\lim_{x\rightarrow 0}\frac{\mathrm{f}(x+a)-\mathrm{f}(a)}{x}=\mathrm{f}^\prime(a)$$
does not count as "using l'Hopital's rule."

4. Jun 1, 2014

### lurflurf

you have a mistake somewhere post more steps if you cannot find it yourself

you also could use
$$2\sin^2(x/2)+1-\cos^2(x)=2\sin^2(x/2)(2+\cos(x))$$