Compute Limit w/ L'Hospital: Is This Right?

[endeavor]

Homework Statement

Compute the limit \lim_{x \rightarrow 0} (1 + 3x)^{3/x}.

Homework Equations

The Attempt at a Solution

\lim_{x \rightarrow 0} (1 + 3x)^{3/x} = \lim_{x \rightarrow 0} e^{\frac{3}{x} \ln (1 + 3x)}
\lim_{x \rightarrow 0} \frac{3}{x} \ln (1 + 3x) = 3 \lim_{x \rightarrow 0} \frac{ln (1 + 3x)}{x}
= 3 \lim_{x \rightarrow 0} \frac{\frac{3}{1 + 3x}}{1}
= 3 \lim_{x \rightarrow 0} \frac{3}{1 + 3x}
=9
\lim_{x \rightarrow 0} (1 + 3x)^{3/x} = e^9

I just need to know if this is right.

 

[Dick]

It is just fine.

 

[IMDerek]

The answer is right. Just plug in .01 or .001 into the limit using the calculator and compare it to the exact value.

 

[Gib Z]

Or You could have manipulated the definition of the exponential: e^a=\lim_{x \rightarrow \inf} (1+\frac{a}{x})^x

Rewrite \lim_{x \rightarrow 0} (1 + 3x)^{3/x} as \lim_{x \rightarrow 0} (1 + \frac{3}{1/x})^{3(1/x)}
The limit of 1/x as 0 approaches zero is the same as x appracohes infinity. So we can rewrite the limit as
\lim_{x \rightarrow \inf} (1 + 3/x)^{3x} = \lim_{x \rightarrow \inf}( (1 + 3/x)^x )^3 which is the definition of e^3, then cubed, getting out e^9 as required.