How Does Differentiability Affect Limits Involving Exponentials?

[snipez90]

Homework Statement

Suppose f is differentiable at a and f(a) =/= 0. Evaluate
\lim_{n \rightarrow \infty}\left[\frac{f(a+\frac{1}{n})}{f(a)}\right]^n

Homework Equations

None really

The Attempt at a Solution

Can someone check my argument:

Since f is differentiable at a, f is continuous at a, so for sufficiently large n, f(a + 1/n) and f(a) have the same sign. Hence
\lim_{n \rightarrow \infty}\left[\frac{f(a+\frac{1}{n})}{f(a)}\right]^n = \lim_{n \rightarrow \infty}\exp{\left(n\log{\left(\frac{|f(a+\frac{1}{n})|}{|f(a)|}\right)}\right)} = \exp{\left(\lim_{n \rightarrow \infty}\frac{\log{|f(a+\frac{1}{n})|} - \log{|f(a)|}}{\frac{1}{n}}\right)}.
This last limit in the argument of the exponential is the derivative of log|f(x)| at x = a, which is f'(a)/f(a). Thus, the original limit is simply e^(f'(a)/f(a)).

 

[VietDao29]

Homework Statement

Suppose f is differentiable at a and f(a) =/= 0. Evaluate
\lim_{n \rightarrow \infty}\left[\frac{f(a+\frac{1}{n})}{f(a)}\right]^n

Homework Equations

None really

The Attempt at a Solution

Can someone check my argument:

Since f is differentiable at a, f is continuous at a, so for sufficiently large n, f(a + 1/n) and f(a) have the same sign. Hence
\lim_{n \rightarrow \infty}\left[\frac{f(a+\frac{1}{n})}{f(a)}\right]^n = \lim_{n \rightarrow \infty}\exp{\left(n\log{\left(\frac{|f(a+\frac{1}{n})|}{|f(a)|}\right)}\right)} = \exp{\left(\lim_{n \rightarrow \infty}\frac{\log{|f(a+\frac{1}{n})|} - \log{|f(a)|}}{\frac{1}{n}}\right)}.
This last limit in the argument of the exponential is the derivative of log|f(x)| at x = a, which is f'(a)/f(a). Thus, the original limit is simply e^(f'(a)/f(a)).

Looks perfect to me. :)