# L'Hopitals Rule

Evaluate limit

lim x--> 2+ (5(x-2))^(x-2)

so what i did i let y= that limit
Then i took the natural log:

lim x --> 2+ (x-2) ln| (5(x-2))

then i put it in the form of L'hopitals rule:

lim x--> 2+ (ln |(5(x-2))) / 1/(x-2) ...when i plug in 2 for x, both equations DNE so it fits the rule

So i get the function [5(5x-10)^-1] / [ -(x-2)^-2]
I just find that by looking at what the equation is, every time i apply l'hopitals rule, it will just keep growing within the negative powers.. im not sure what to do, if i did something wrong, or what i should try?

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Dick
Homework Helper
Try and simplify the algebra before you do l'Hopital again.

simplify it anymore? like square the denominator?:

[5/(5x-10)] / [-1/ (x^2-4x+4)]
thats all i can think off...

vela
Staff Emeritus
Homework Helper
You just need to simplify the expression you got after applying L'Hopital's rule, and you'll see nothing blows up when you let x go to 2.

Dick
Homework Helper
(x-2) divides (5x-10). There's a more profitable way to simplify.

oh wait... would i re-raise the equation from the denominator and like re-due l'hopitals rule: here is what i mean if that was confusing

Here was the equation we had when i first used the rule:
[5(5x-10)^-1] / [ -(x-2)^-2]

now ill re raise it
-(x-2)^2* [5/ (5x-10)]

and reuse l'hopitals rule but with new equations in the numerator and denominator:
[ -5(x-2)^2] / [1/ (5x-10)]

i dont know... maybe im going in the wrong direction.

Dick
Homework Helper
oh wait... would i re-raise the equation from the denominator and like re-due l'hopitals rule: here is what i mean if that was confusing

Here was the equation we had when i first used the rule:
[5(5x-10)^-1] / [ -(x-2)^-2]

now ill re raise it
-(x-2)^2* [5/ (5x-10)]

and reuse l'hopitals rule but with new equations in the numerator and denominator:
[ -5(x-2)^2] / [1/ (5x-10)]

i dont know... maybe im going in the wrong direction.
Cancel the common factor before you do anything. And your algebra doesn't quite look right there.

sorry the new equation would be
[ -5(x-2)^2] / [5x-10]

Dick
Homework Helper
You can l'Hopital it again now, or you can just cancel the common factor.

Oh i think i understand what your saying
the equation would be:
5/ (5x-10) * (x-2)^2

you can simplify the 5x-10 by doing (5(x-2)) So..

5/5(x-2) * x-2^2

so the x-2 gets rid off and the sqrd on top gets gone leaving:

5/5 * x-2 or simply x-2?

sorry i forgot to include the negative:
so final answer would be -(x-2) or (2-x)

so the limit as x--> 2+ for the ln of the function is 0?

vela
Staff Emeritus