Finding the Limit of x->0+ sqrtx/(1-cosx)

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Discussion Overview

The discussion centers around finding the limit of the expression sqrt(x)/(1-cos(x)) as x approaches 0 from the positive side. Participants explore various methods, including L'Hôpital's rule, asymptotic expansions, and algebraic manipulations, while addressing the complexities of the limit's behavior.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant attempts to apply L'Hôpital's rule and arrives at the limit of 1/(2sqrt(x)sin(x)), expressing difficulty in progressing further.
  • Another participant questions the behavior of sqrt(x) and sin(x) as x approaches 0 from the positive side.
  • A suggestion is made to use the asymptotic expansion of cosine, specifically that cos(x) approximates to 1 - (x^2)/2 when x is near zero.
  • Some participants express confusion about the form of the limit, with one noting it appears to be an indeterminate form of infinity * infinity.
  • Another participant suggests rewriting the expression to facilitate finding the limit, indicating a potential simplification.
  • There is a mention of a rule related to limits where the numerator approaches 0 while the denominator approaches a non-zero value, with a participant asserting the limit is +0.
  • One participant graphs the functions involved and concludes that sqrt(x) grows larger than 1-cos(x) as x approaches 0, suggesting the limit might be infinity.
  • Another participant acknowledges a mistake in their earlier reasoning, realizing the need to apply L'Hôpital's rule again.
  • There is a proposal to factor the expression into two parts, both of which approach infinity, suggesting a path to evaluate the limit.

Areas of Agreement / Disagreement

Participants express differing views on the limit's behavior, with some suggesting it approaches 0, while others propose it approaches infinity. The discussion remains unresolved, with multiple competing perspectives presented.

Contextual Notes

Participants reference various mathematical rules and forms of indeterminacy, indicating a reliance on specific limit properties that may not be universally agreed upon. There are also indications of confusion regarding the application of L'Hôpital's rule and the behavior of the functions involved near the limit.

apchemstudent
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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.
 
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apchemstudent said:
lim
x->0+ 1/(2sqrtx*sinx)

What's sqrt(x) going to as x->0+? sin(x)?
 
Use the asymptotic expansion of cosine

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

Daniel.
 
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.
 
Check your list of indeterminate forms.
 
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
 
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..
 
No you don't have to make it hard..beleive me it so simple...we did in calc 1 I forgot the rule name.
 
So, why is your "answer" dead wrong, then? :wink:
 
  • #10
opps didnt noticed bottom was going to 0 too...haha
then you will have to use l'hopital rule.
 
  • #11
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
no i seem to be right, i.e. one use of l'hopital gives it, yes?
 
  • #13
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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