Doubt about two limits (short)

[Felafel]

Homework Statement

Find the limit of
$1): \displaystyle \lim_{n \to +\infty}(\frac{f(a+\frac{1}{n})}{f(a)})^{\frac{1}{n}}$
$2) \displaystyle \lim_{x \to a} (\frac{f(x)}{f(a)})^{\frac{1}{ln(x)-ln(a)}}(=1^{\infty})$
I am not quite sure if i can solve it the way I did, it has been to easy so there's a trick, I'm afraid

The Attempt at a Solution

$1) \displaystyle \lim_{n \to +\infty}(\frac{f(a+\frac{1}{n})}{f(a)})^{\frac{1}{n}}$
=$\displaystyle \lim_{n \to +\infty} e^{\frac{1}{n} (log(f(a+\frac{1}{n})-log(f(a)))}$=
=$e^0=1$

$2) \displaystyle \lim_{x \to a} (\frac{f(x)}{f(a)})^{\frac{1}{ln(x)-ln(a)}}(=1^{\infty})$
$\displaystyle \lim_{x \to a} e^{ln(\frac{f(x)}{f(a)}) \frac{1}{ln(x)- ln(a)}}$
$\displaystyle \lim_{x \to a} \frac{ln(f(x))-ln(f(a))}{ln(x)- ln(a)}(=\frac{0}{0})$ with de l'Hopital
$\displaystyle \lim_{x \to a} \frac{f'(x)}{xf(x)}=\frac{f'(a)}{af(a)}$

 

[SammyS]

Homework Statement

Find the limit of
$1): \displaystyle \lim_{n \to +\infty}(\frac{f(a+\frac{1}{n})}{f(a)})^{\frac{1}{n}}$
$2) \displaystyle \lim_{x \to a} (\frac{f(x)}{f(a)})^{\frac{1}{ln(x)-ln(a)}}(=1^{\infty})$
I am not quite sure if i can solve it the way I did, it has been too easy so there's a trick, I'm afraid

The Attempt at a Solution

$1) \displaystyle \lim_{n \to +\infty}(\frac{f(a+\frac{1}{n})}{f(a)})^{\frac{1}{n}}$
=$\displaystyle \lim_{n \to +\infty} e^{\frac{1}{n} (log(f(a+\frac{1}{n})-log(f(a)))}$=
=$e^0=1$

$2) \displaystyle \lim_{x \to a} (\frac{f(x)}{f(a)})^{\frac{1}{ln(x)-ln(a)}}(=1^{\infty})$
$\displaystyle \lim_{x \to a} e^{ln(\frac{f(x)}{f(a)}) \frac{1}{ln(x)- ln(a)}}$
$\displaystyle \lim_{x \to a} \frac{ln(f(x))-ln(f(a))}{ln(x)- ln(a)}(=\frac{0}{0})$ with de l'Hopital
$\displaystyle \lim_{x \to a} \frac{f'(x)}{xf(x)}=\frac{f'(a)}{af(a)}$

The derivative of ln(x) is 1/x.

You have 1/x in the denominator. That becomes x in the numerator. (You could have checked your result with an appropriate example.)

 

[Ray Vickson]

Homework Statement

Find the limit of
$1): \displaystyle \lim_{n \to +\infty}(\frac{f(a+\frac{1}{n})}{f(a)})^{\frac{1}{n}}$
$2) \displaystyle \lim_{x \to a} (\frac{f(x)}{f(a)})^{\frac{1}{ln(x)-ln(a)}}(=1^{\infty})$
I am not quite sure if i can solve it the way I did, it has been to easy so there's a trick, I'm afraid

The Attempt at a Solution

$1) \displaystyle \lim_{n \to +\infty}(\frac{f(a+\frac{1}{n})}{f(a)})^{\frac{1}{n}}$
=$\displaystyle \lim_{n \to +\infty} e^{\frac{1}{n} (log(f(a+\frac{1}{n})-log(f(a)))}$=
=$e^0=1$

$2) \displaystyle \lim_{x \to a} (\frac{f(x)}{f(a)})^{\frac{1}{ln(x)-ln(a)}}(=1^{\infty})$
$\displaystyle \lim_{x \to a} e^{ln(\frac{f(x)}{f(a)}) \frac{1}{ln(x)- ln(a)}}$
$\displaystyle \lim_{x \to a} \frac{ln(f(x))-ln(f(a))}{ln(x)- ln(a)}(=\frac{0}{0})$ with de l'Hopital
$\displaystyle \lim_{x \to a} \frac{f'(x)}{xf(x)}=\frac{f'(a)}{af(a)}$

Note: that should be l'Hospital"---like the place with the emergency ward---not "l'Hopital" (although it is pronounced as "lo-pee-tal", with a silent 's').

 

[Whovian]

Note: that should be l'Hospital"---like the place with the emergency ward---not "l'Hopital" (although it is pronounced as "lo-pee-tal", with a silent 's').

I tend to see it as l'Hôpital, so maybe they read l'Hôpital and didn't know how to type an ô?

 

[Felafel]

thanks everyone!