L'Hopitals rule and application in limits and limits

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Homework Statement


http://img227.imageshack.us/img227/9792/hopitalsruleandlimits.png [Broken]
Ive added a picture of the question. I know that L'hopitals rule is lim x->a f(x)/g(x) = lim x->a f'(x)/g'(x) but I dont understand how to apply it in this situation >< and the other questions I dont understand how to start off and what to get to. If anyone could please help me with ANY of the questions i would be so grateful!

Homework Equations




The Attempt at a Solution


progress so far
a) I don't know how get it from positive and from the negative side using the rule, but I used values and from below 1/2 and it approaches 0 from the positive and from above 1/2 it approaches 0

b) i) I've shown its an even function using x= 1 and x= -1
ii) uses stuff from a) so I dont know how to do it. but i assume it does has a limit.. i dont know an explanation though
iii) wild guess.. limit at -1/2 and 1/2 should be.. 0? but why? (if it does = 0)

c) i) I know how to find the derivative using the product rule but I dont understand why its x > 1/2 and -1/2 < x < 1/2
ii) similar to a) dont know how to do that
iii) dont know
iv) f'(3/4) i subbed in the value to upper derivative from above and i get to log [ (5/4)1/4 / 45/4 ] and im not sure how to go from there
f'(5/6) = log [ (4/3)1/3 / 34/3 how can i simplify that..?
and i dont know what the second part of the question means
v) dont know

d) any easy way of finding f"(x)? or do i have to find the equ using quotient rule and not sure what the explanation should be

e) same with a), not sure how to calculate from infinity from +side and -side.. sub values?
 
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Answers and Replies

  • #2
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Your link does not work
 
  • #3
gabbagabbahey
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a) I don't know how get it from positive and from the negative side using the rule, but I used values and from below 1/2 and it approaches 0 from the positive and from above 1/2 it approaches 0

Right, the limit will indeed be zero from both sides. To show this explicitly, use the hint they provided.

For [itex]x>\frac{1}{2}[/itex],

[tex]\begin{aligned}\ln\left|x-\frac{1}{2}\right|\ln\left|x+\frac{1}{2}\right| &= \ln\left(x-\frac{1}{2}\right)\ln\left(x+\frac{1}{2}\right) \\ &=\left(\frac{\ln\left(x-\frac{1}{2}\right)}{\frac{1}{x-\frac{1}{2}}}\right)\left(\frac{\ln\left(x-\frac{1}{2}\right)}{x-\frac{1}{2}}\right) \end{aligned}[/tex]

If you simply substitute [itex]x=\frac{1}{2}[/itex] into this expression, you get [itex]\left(\frac{\infty}{\infty}\right)\left(\frac{0}{0}\right)[/itex] and so it is in one of the forms that allows you to use L'Hopitals rule.

b) i) I've shown its an even function using x= 1 and x= -1

But that doesn't show its even...it only shows that [itex]f(1)=f(-1)[/itex]...to show its even for all [itex]x[/itex] (except [itex]x=\pm\frac{1}{2}[/itex]), you need to show that [itex]\ln\left|(-x)-\frac{1}{2}\right|\ln\left|(-x)+\frac{1}{2}\right|=\ln\left|x-\frac{1}{2}\right|\ln\left|x+\frac{1}{2}\right|[/itex]. That should be fairly easy because of the absolute values (use the fact that [itex]|-a|=|a|[/itex]).

ii) uses stuff from a) so I dont know how to do it. but i assume it does has a limit.. i dont know an explanation though

In part [itex]a[/itex], you should end up with [itex]\lim_{x\to \frac{1}{2}}f(x)=0[/itex]...using the substitution [itex]u=-x[/itex], you can write this as

[tex]\lim_{(-u)\to \frac{1}{2}}f(-u)=\lim_{u\to -\frac{1}{2}}f(-u)=0[/tex]

If [itex]f(x)[/itex] is even, what does that tell you? (keep in mind, that in this limit, [itex]u[/itex] is essentially a dummy variable and can be replaced by [itex]x[/itex], which should make the result easier to see)
 

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