Homework Help: 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