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Indeterminate limit problem

  • Thread starter khwarizm
  • Start date
  • #1
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Homework Statement


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 [itex]\displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}{|x|}}[/itex] 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:
[itex]
\displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1-cos^2(x)}}{|x|}}
[/itex]


The Attempt at a Solution


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

Answers and Replies

  • #2
pasmith
Homework Helper
1,749
421

Homework Statement


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 [itex]\displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}{|x|}}[/itex] 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:
[itex]
\displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1-cos^2(x)}}{|x|}}
[/itex]


The Attempt at a Solution


this is the last step that i came.
[itex]
\displaystyle\lim_{x\rightarrow 0^+} {\frac{1-cos(x)+1-cos^{2}(x)}{|x|\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}}
[/itex]
Does recognising that if [itex]f(0) = 0[/itex] then [tex]
f'(0) = \lim_{x \to 0^{+}} \frac{f(x)-f(0)}{x} = \lim_{x \to 0^{+}} \frac{f(x)}{|x|}
[/tex]count as "using l'Hopital's rule"?

I personally would use [itex]\cos^2(x/2) - \sin^2(x/2) = \cos \theta[/itex] to express [itex]2\sin^{2}(x/2)+1+\cos^2(x)[/itex] as a quadratic in [itex]\cos x[/itex] and then complete the square before differentiating.
 
  • #3
lurflurf
Homework Helper
2,432
132
$$\frac{\sqrt{2\sin^{2}(x/2)+1-\cos^2(x)}}{|x|}=\sqrt{\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2+\left(\frac{sin(x)}{x}\right)^2}$$
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
lurflurf
Homework Helper
2,432
132
you have a mistake somewhere post more steps if you cannot find it yourself

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

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