# Limits w/ l'hospital

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

2. Homework 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.

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## Answers and Replies

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Dick
Homework Helper
It is just fine.

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
Homework Helper
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.

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