What is the indeterminate limit problem?

[khwarizm]

Homework Statement

i need to find the value of limit but its indeterminate and i can't 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.

Homework Equations

This is the question:
<br /> \displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1-cos^2(x)}}{|x|}}<br />

The Attempt at a Solution

this is the last step that i came.
<br /> \displaystyle\lim_{x\rightarrow 0^+} {\frac{1-cos(x)+1-cos^{2}(x)}{|x|\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}}<br />

 

[pasmith]

Homework Statement

i need to find the value of limit but its indeterminate and i can't 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.

Homework Equations

This is the question:
<br /> \displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1-cos^2(x)}}{|x|}}<br />

The Attempt at a Solution

this is the last step that i came.
<br /> \displaystyle\lim_{x\rightarrow 0^+} {\frac{1-cos(x)+1-cos^{2}(x)}{|x|\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}}<br />

Does recognising that if f(0) = 0 then <br /> f&#039;(0) = \lim_{x \to 0^{+}} \frac{f(x)-f(0)}{x} = \lim_{x \to 0^{+}} \frac{f(x)}{|x|}<br />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.

 

[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."

 

[lurflurf]

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

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