L'Hopitals rule and application in limits and limits

In summary: From what you've shown, it tells you that for all x except possibly x=\pm\frac{1}{2}, f(x) is equal to \ln(x-\frac{1}{2}).
  • #1
moocav
6
0

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 don't understand how to apply it in this situation >< and the other questions I don't 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 don't know how to do it. but i assume it does has a limit.. i don't 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 don't understand why its x > 1/2 and -1/2 < x < 1/2
ii) similar to a) don't know how to do that
iii) don't 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 I am not sure how to go from there
f'(5/6) = log [ (4/3)1/3 / 34/3 how can i simplify that..?
and i don't know what the second part of the question means
v) don't 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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  • #2
Your link does not work
 
  • #3
moocav said:
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 don't know how to do it. but i assume it does has a limit.. i don't 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)
 

What is L'Hopital's rule?

L'Hopital's rule, also known as the rule of de L'Hopital, is a mathematical theorem used to evaluate limits involving indeterminate forms such as 0/0 or ∞/∞. It states that for two functions f(x) and g(x), if their derivative approaches the same limit at a given point, then the limit of their quotient will be the same as well.

When can L'Hopital's rule be applied?

L'Hopital's rule can only be applied when the limit of a function is in an indeterminate form. This means that when directly evaluating the limit results in an undefined value or does not provide any information about the behavior of the function at that point.

What is the process for using L'Hopital's rule?

The process for applying L'Hopital's rule involves taking the derivative of both the numerator and denominator of the original function and then evaluating the limit using the new derivatives. If the limit is still in an indeterminate form, the process can be repeated until a definitive value is obtained.

Are there any limitations to using L'Hopital's rule?

Yes, there are limitations to using L'Hopital's rule. It can only be applied to certain types of limits, such as those involving indeterminate forms, and it may not always provide a definitive answer. It is important to also consider other methods for evaluating limits and to use L'Hopital's rule as a supplement rather than the sole method.

Can L'Hopital's rule be applied to limits approaching infinity?

Yes, L'Hopital's rule can also be applied to limits approaching infinity. In these cases, the limit can be rewritten as a quotient of two functions and then L'Hopital's rule can be used to evaluate the limit. However, it is important to consider the behavior of the functions as x approaches infinity and to use L'Hopital's rule as a supplement rather than the sole method for evaluating the limit.

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