# Need a hint on this limit

1. Jun 15, 2005

### jdstokes

$\lim_{x\rightarrow 0^+}(1+3x)^{1/x}$

Thanks.

2. Jun 15, 2005

### steven187

hello there

firstly i would make a substitution such as
$x=\frac{1}{u}$
$\lim_{x\rightarrow 0^+}(1+3x)^{1/x}=\lim_{u\rightarrow\infty}(1+\frac{3}{u})^{u}$ you do know what this limit is equal to right?

steven

3. Jun 15, 2005

### HallsofIvy

Staff Emeritus
It might be better to make the substitution $$x= \frac{1}{3u}$$ so that $$1+ 3x= 1+ \frac{1}{u}$$ and $$\frac{1}{x}= 3u$$. Then the limit becomes
$$\lim_{x\rightarrow 0^+}(1+3x)^{1/x}=\lim_{u\rightarrow\infty}(1+\frac{1}{u})^{3u}= \{\lim_{u\rightarrow\infty}(1+ \frac{1}{u})^u\}^3$$.

Last edited: Jun 15, 2005
4. Jun 15, 2005

### jdstokes

I still don't know where to go with this. I seem to keep getting a $(1+ 0)^\infty$ situation.

5. Jun 15, 2005

### dextercioby

Do you know the definition of "e" (Euler's number)...?

Daniel.

6. Jun 15, 2005