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apoptosis

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Heya! Here's a question that I CANNOT solve for the life of me. Thanks for any direction!

lim (lnx)[tex]^{\frac{1}{x-e}}[/tex]

x[tex]\rightarrow[/tex]e+

^^ I'm not sure if Latext is clear, but x-e is an exponent to lnx

So, I'm thinking L'hopital's rule in starting off the question, but I'm not exactly sure if it's the right way. my work so far is in part 3...

To bring the exponent down, I bring in ln:

ln(lnx)[tex]\frac{1}{x-e}[/tex]

so: [tex]\frac{ln(lnx)}{x-e}[/tex]

applying l'hopitals, I find that it results in a 0/0 fraction so I derive:

to get:

[tex]\frac{1}{x)(lnx)(1-e)}[/tex]

However, when I sub in e I get -2.14...so I guess I'm on the wrong track

I did graph the function, where the equation is not defined at e. As I approach e from the right, I get a number of 1.44.

Thank you once again!

## Homework Statement

lim (lnx)[tex]^{\frac{1}{x-e}}[/tex]

x[tex]\rightarrow[/tex]e+

^^ I'm not sure if Latext is clear, but x-e is an exponent to lnx

## Homework Equations

So, I'm thinking L'hopital's rule in starting off the question, but I'm not exactly sure if it's the right way. my work so far is in part 3...

## The Attempt at a Solution

To bring the exponent down, I bring in ln:

ln(lnx)[tex]\frac{1}{x-e}[/tex]

so: [tex]\frac{ln(lnx)}{x-e}[/tex]

applying l'hopitals, I find that it results in a 0/0 fraction so I derive:

to get:

[tex]\frac{1}{x)(lnx)(1-e)}[/tex]

However, when I sub in e I get -2.14...so I guess I'm on the wrong track

I did graph the function, where the equation is not defined at e. As I approach e from the right, I get a number of 1.44.

Thank you once again!

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