Solving Limit Problems with L'Hopital's Rule

[rdougie]

Homework Statement

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

(b) Find the limit.

Homework Equations

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

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

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

lim$$_{x\rightarrow0}$$sinx/x = 1



3. My attempt(s) at a solution

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

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

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

 

[NeoDevin]

Try thinking of the original problem as

$$\lim_{x\rightarrow 0} \frac{x\sin(x)}{\frac{\sin(x)}{x}}$$

 

[rdougie]

Where did sin(1/x) go?

 

[NeoDevin]

Sorry

$$\lim_{x\rightarrow 0} \frac{x\sin\left(\frac{1}{x}\right)}{\frac{\sin(x)}{x}}$$

 

[rdougie]

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?

 

[NeoDevin]

the limit is zero for the numerator, but check the list you posted in the first post again, and see what the limit for $\frac{\sin(x)}{x}$ is.

 

[rdougie]

ooh sorry lim sinx/x = 1.
ok thx!

 

[HallsofIvy]

So far you haven't said anything about
(a) Explain why L'Hopital's rule does not apply to the problem
lim_x->0 [ (xsin(1/x)) / sinx ]

 

[NeoDevin]

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 $\frac{x}{x}$, or $\frac{e^{-1/x}}{e^{-1/x}}$ 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.

 

[ZioX]

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.

 

[rdougie]

So far you haven't said anything about
(a) Explain why L'Hopital's rule does not apply to the problem
lim_x->0 [ (x$$^{2}$$sin(1/x)) / sinx ]

lim x^2 = 0; lim sin(1/x) DNE; lim sinx = 0; 0 * DNE / 0 isn't one of the indeterminant forms.

thanks for the help, everbody :)

 

[tommygun101]

hey,..

l' hopitals rule doesn't 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 don't 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

 

[ZioX]

lim x^2 = 0; lim sin(1/x) DNE; lim sinx = 0; 0 * DNE / 0 isn't 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?