Limit problem

  • #1

Main Question or Discussion Point

How do you find the limit to:

Lim
x->0+ sqrtx/(1-cosx)

I use the L'Hopital rule and i get

lim
x->0+ 1/(2sqrtx*sinx)

and I can't get anywhere after this. I even tried multiplying by sqrtx/sqrtx to get rid of the sqrtx at the beginning, but it still doesn't work. Please help, thanks.
 

Answers and Replies

  • #2
shmoe
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apchemstudent said:
lim
x->0+ 1/(2sqrtx*sinx)
What's sqrt(x) going to as x->0+? sin(x)?
 
  • #3
dextercioby
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Use the asymptotic expansion of cosine

[tex] \cos x \simeq 1-\frac{x^{2}}{2} [/tex] when "x" is close to zero.

Daniel.
 
  • #4
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it's still in the form infinity * infinity isn't it? got to use l'hopital's rule one more time. i think the answer is 0.
 
  • #5
Hurkyl
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Check your list of indeterminate forms.
 
  • #6
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it is simple I guess...use the
"when top is going to 0 and bottom is going to soem where else."......I forgot the rule...But the answer is +0
 
  • #7
arildno
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Just rewrite your expression as
[tex]\frac{\sqrt{x}}{1-\cos(x)}=\frac{1}{x^{\frac{3}{2}}}(\frac{x}{\sin(x)})^{2}(1+\cos(x))[/tex]
Then it is easy to see what you should get.
Alternatively, use dexter's approach..
 
  • #8
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No you dont have to make it hard..beleive me it so simple....we did in calc 1 I forgot the rule name.
 
  • #9
arildno
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So, why is your "answer" dead wrong, then? :wink:
 
  • #10
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opps didnt noticed bottom was going to 0 too...haha
then you will have to use l'hopital rule.
 
  • #11
mathwonk
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well i just graphed them and looked at the graphs. sqrt(x) is really big compared to x, while 1-cos(x) is really small, so I think the limit is infinity. but now i will do the math, and no doubt come back chagrined.
 
  • #12
mathwonk
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no i seem to be right, i.e. one use of l'hopital gives it, yes?
 
  • #13
mathwonk
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or just write it as (sqrt(x)/x). (x/[1-cos(x))]) and both factors go to + infinity, the first trivially and the second by l'hopital.
 

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