# Calculus

posted by .

L'Hospital's Rule

Lim(lnx)^(x-1)
x->1+

This is what I did so far...
Form: 0^0
ln(lnx)/ (1/x-1)

-----------------------------------

I am not sure if you could use L'Hospital Rule (as stated in the instruction)for the 2 problems below:

lim [ln(y^2+2y)]/[lny]=1
y->0+

lim xlnx = 0
x->0+

I was able to solve them without using the Rule.

• Calculus -

Actually, I would not recommend using any rules unless you really understand why and how they work. I.e. you should be able to derive those rules from first principles. Otherwise you are just training to become some sort of math computer who does not really understand math :)

L'Hopital's rule is actually closely related to Taylor's theorem which says that foir a sufficiently differentiable function f one has:

f(x+h) = f(x) + h f'(x) + h^2/2 f''(x) + h^3/6 f'''(x) + higher order terms in h.

This is something that you can easily derive and intuitively understand. Near a point x the function f will be f(x) plus correction terms. You can apply this to compute the limits as follows:

In case of (lnx)^(x-1) you take the logarithm, using that the log of the limit is the limit of the log which follows from the fact the the logarithm is a continuous function.

You then get the function

(x-1)ln(ln(x))

of which you want to compute the limit x--->1

You then put x = 1 + h and do a series expansion in powers of h. You get:

h(ln(ln(1+h)))

ln(1 + h) = h - h^2/2 + O(h^3)

where O(h^n) means a term proportional to h^n for small h. Taking the log again and using the same series expansion gives:

ln(ln(1 + h)) =
ln(h - h^2/2 + O(h^3)) =

ln(h) + ln(1 - h/2 +O(h^2)) =

ln(h) - h/2 + O(h^2)

Multiplying this by h and taking the limit h -->0 gives -1/2, so the original limit is exp(-1/2)

Note that we needed to go to second order in the Taylor expansion, which means that if you had used L'Hopital you would have had to use that rule twice.

• Calculus: Correction -

Sorry, I was a bit confused, when multiplying by h the limit becomes zero so after exponentiationg you get 1. So, You could have used L'Hopital's rule once...

• Calculus -

im going to figure this one out!

## Similar Questions

1. ### Calculus

Below are the 5 problems which I had trouble in. I can't seem to get the answer in the back of the book. Thanks for the help! lim (theta-pi/2)sec(theta) theta->pi/2 Answer: -1 I am not sure what to do here. lim (tan(theta))^(theta) …
2. ### calculus

The limit as x approaches infinity of (e^x+x)^(1/x). I got that it diverges, but I'm not sure if I made a mistake. My work: lim(e+x^(1/x)) lim(e+(1/x^x)) lim(ex^x+1)/x^x l'hopital:lim(e^(e(lnx+1))+1)/e^(lnx+1) diverges?
3. ### Calculus

Find the derivative of y with respect to x. y= [(x^5)/5]lnx-[(x^5)/25] How would I go about doing this?
4. ### L'Hopital's rule

Find lim x->1+ of [(1/(x-1))-(1/lnx)]. Here is my work... =(lnx-(x-1)) / ((x-1)(lnx)) =(lnx-1) / (lnx+ (x+1)/x) This becomes(1/x) / ((1/x)+(1/x^2)) which becomes 1/ (1/x^2) This equals 1/2. I understand the answer has to be -1/2, …
5. ### calculus

Evaluate the following limits after identifying the indeterminate form. Use Hospital's rule. d) lim x_0+ (xe^(2x) + 1)^(5/x) e) lim x_(Pi/2)+ (1 + sec3x)^(cot3x) Thank you!
6. ### Calculus 1

Evaluate the following limits after identifying the indeterminate form. Use Hospital's rule. d) lim x_0+ (xe^(2x) + 1)^(5/x) e) lim x_(Pi/2)+ (1 + sec3x)^(cot3x) Thank you!
7. ### College Calc

Problem: y=sqrtx ^1/x. Find DY/DX. ln(f(x))=lnx^1/2x or lnx to the 1/2x 1/2x lnx product rule (1/4x^2)(lnx)+(1/x)(1/2x) is the product rule correct?
8. ### Calculus 1

Find the derivative of y with respect to x. y=(x^6/6)(lnx)-(x^6/36) So far this is what I've gotten: y=(x^6/6)(lnx)-(x^6/36) y=(1/6)x^6(lnx)-(1/36)x^6 y'=(1/6)x^5(1/x)+lnx(x^5)-(1/6)x^5 What do I do now?
9. ### Calculus Help Please Urgent!!!

find the limit algebraically. Use L'Hospital's Rule where appropriate. If there is a moare elementary method, consider using it. If L'Hospital's Rule doesn't apply, explain why lim -> 0 (cot(x)-(1/x)) show work please!!!
10. ### Calculus

Find the limit as x approaches 0+ of (lnx)/x using L'hospitals rule. When I do this, I keep getting stuck at 1/0 when you plug back into the equation after doing l'hospital once.

More Similar Questions