L'hopital's rule

  • Thread starter rdougie
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  • #1
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Homework Statement



(a) Explain why L'Hopital's rule does not apply to the problem
lim[tex]_{x\rightarrow0}[/tex] [ (x[tex]^{2}[/tex]sin(1/x)) / sinx ]

(b) Find the limit.


Homework Equations



lim [tex]_{x\rightarrow0}[/tex] xsin(1/x) = 0 , by the Squeezing Theorem.

lim [tex]_{x\rightarrow0}[/tex] sin (1/x) Does Not Exist because it oscillates between -1 and 1.

lim [tex]_{x\rightarrow0}[/tex] x[tex]^{2}[/tex]sin(1/x) = 0 by the Squeezing Theorem.

lim[tex]_{x\rightarrow0}[/tex]sinx/x = 1



3. My attempt(s) at a solution

I wrote the original problem
lim[tex]_{x\rightarrow0}[/tex] [ (x[tex]^{2}[/tex]sin(1/x)) / sinx ]

as
lim[tex]_{x\rightarrow0}[/tex] sin (1/x) / lim[tex]_{x\rightarrow}[/tex](1/x) * lim[tex]_{x\rightarrow0}[/tex] (sinx/x).

Since the limit of the numerator doesn't exist, and lim[tex]_{x\rightarrow0}[/tex](1/x) is +[tex]\infty[/tex], and lim[tex]_{x\rightarrow0}[/tex] sinx/x = 1, then the limit of the problem doesn't exist, right?
 

Answers and Replies

  • #2
306
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Try thinking of the original problem as

[tex]\lim_{x\rightarrow 0} \frac{x\sin(x)}{\frac{\sin(x)}{x}}[/tex]
 
  • #3
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Where did sin(1/x) go?
 
  • #4
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Sorry

[tex]\lim_{x\rightarrow 0} \frac{x\sin\left(\frac{1}{x}\right)}{\frac{\sin(x)}{x}}[/tex]
 
  • #5
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So now the limit is 0 if I use the Squeezing Theorem for the numerator, and the "lim sinx/x=0" for the denominator? Am I thinking about this correctly or am I just trying to plug and play?
 
  • #6
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the limit is zero for the numerator, but check the list you posted in the first post again, and see what the limit for [itex]\frac{\sin(x)}{x}[/itex] is.
 
  • #7
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ooh sorry lim sinx/x = 1.
ok thx!
 
  • #8
HallsofIvy
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So far you haven't said anything about
(a) Explain why L'Hopital's rule does not apply to the problem
limx->0 [ (xsin(1/x)) / sinx ]
 
  • #9
306
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ooh sorry lim sinx/x = 1.
ok thx!

I think you usually use L'Hopital's rule to get that result in the first place though.

This seems like a poorly though out question to me, as any expression could be modified to make use of L'Hopitals rule, by multiplying by [itex]\frac{x}{x}[/itex], or [itex]\frac{e^{-1/x}}{e^{-1/x}}[/itex] Then taking the derivative will give you the original result.

In answer to the question a, I would say that L'hopital's rule applies, either to the expression itself as it's written, or to the denominator when you rewrite it.
 
  • #10
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L'hopitals rule DOES apply (okay maybe it doesn't, but it would, if it could)! The numerator is defined on R except at 0. The denominator is defined on all of R. Both are differentiable on their respective natural domains. The derivative of the denominator is nonzero in a deleted nbd of 0. Does limit of the quotient of the derivatives exist though?

You'll want to use the sequential criterion to show that the limit does not exist.
 
Last edited:
  • #11
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So far you haven't said anything about
(a) Explain why L'Hopital's rule does not apply to the problem
limx->0 [ (x[tex]^{2}[/tex]sin(1/x)) / sinx ]

lim x^2 = 0; lim sin(1/x) DNE; lim sinx = 0; 0 * DNE / 0 isnt one of the indeterminant forms.

thanks for the help, everbody :)
 
  • #12
hey,..

l' hopitals rule doesnt apply here because sin(1/x) and cos(1/x) oscilate rapidly at x near zero 1/x is a simple pole at x = 0,... and derivatives of all orders dont get rid of the sin(1/x) and higher derivatives give the term 1/x^n which tends to infinity as x -> 0.

at least i think so
 
  • #13
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lim x^2 = 0; lim sin(1/x) DNE; lim sinx = 0; 0 * DNE / 0 isnt one of the indeterminant forms.

thanks for the help, everbody :)

Not quite. x^2sin(1/x) -> 0 (product of bounded function and one going to zero) so we have 0/0 which is one of the indeterminate forms. Did you even read anything I wrote previously?
 

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