# Limits w/ l'hospital

1. Jan 27, 2007

### endeavor

1. The problem statement, all variables and given/known data
Compute the limit $$\lim_{x \rightarrow 0} (1 + 3x)^{3/x}$$.

2. Relevant equations
3. 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.

Last edited: Jan 27, 2007
2. Jan 27, 2007

### Dick

It is just fine.

3. Jan 27, 2007

### IMDerek

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

4. Jan 27, 2007

### 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 definiton of e^3, then cubed, getting out e^9 as required.

Last edited: Jan 27, 2007