# Evaluating Limit: lim x--> 2+ (5(x-2))^(x-2)

• mattmannmf

#### mattmannmf

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.. I am not sure what to do, if i did something wrong, or what i should try?

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...

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.

(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 don't know... maybe I am going in the wrong direction.

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 don't know... maybe I am 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]

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?

Yup.