1. L'Hospital's Rule

Use L'Hospital's Rule to find the following limit:

$\displaystyle \lim_{x\rightarrow\infty} (1+\frac {2}{x})^{x}$

I am not sure how to use L'Hospital's rule for this.

I know that the $\displaystyle \lim_{x\rightarrow\infty} \frac{2}{x} = 0$

and so therefore,

$\displaystyle \lim_{x\rightarrow\infty} (1 + 0)^{x} = 1^{\infty} = 1$

Can someone help me solve this using L'Hospital's Rule.

Thank you.

2. Originally Posted by mybrohshi5
Use L'Hospital's Rule to find the following limit:

$\displaystyle \lim_{x\rightarrow\infty} (1+\frac {2}{x})^{x}$

I am not sure how to use L'Hospital's rule for this.

I know that the $\displaystyle \lim_{x\rightarrow\infty} \frac{2}{x} = 0$

and so therefore,

$\displaystyle \lim_{x\rightarrow\infty} (1 + 0)^{x} = 1^{\infty} = 1$

Can someone help me solve this using L'Hospital's Rule.

Thank you.
$\displaystyle \lim_{x\to \infty }\left (1+\frac{2}{x} \right )$$\displaystyle =\lim_{x\to \infty }\left (\frac{x+2}{x} \right )$$\displaystyle =\lim_{x\to \infty }(\frac{1}{1})=1$ (by the Hospital's rule).
therefore $\displaystyle \lim_{x\to \infty }(1)^x=1$

3. That's wrong, Raoh :P

$\displaystyle \lim_{x\to \infty }\left (1+\frac{1}{x} \right ) = \lim_{x\to \infty }\left (\frac{x+1}{x} \right ) = 1$, but
$\displaystyle \lim_{x\to \infty }\left (1+\frac{1}{x} \right )^x = e$

What you need to do here:
Set $\displaystyle t = \left (1+\frac{2}{x} \right )^x$, evaluate $\displaystyle ln(t)$ and take the limit as x goes to infinity. (Hint: the answer is not 1)

4. Originally Posted by Defunkt
That's wrong, Raoh :P

$\displaystyle \lim_{x\to \infty }\left (1+\frac{1}{x} \right ) = \lim_{x\to \infty }\left (\frac{x+1}{x} \right ) = 1$, but
$\displaystyle \lim_{x\to \infty }\left (1+\frac{1}{x} \right )^x = e$

What you need to do here:
Set $\displaystyle t = \left (1+\frac{2}{x} \right )^x$, evaluate $\displaystyle ln(t)$ and take the limit as x goes to infinity. (Hint: the answer is not 1)
errr !
$\displaystyle \ln(t)=x\ln(1+\frac{2}{x})$
$\displaystyle \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x}) \right )=2$.
hope it's right now

5. Im not exactly sure how by setting t equal to the function then taking the natural log of it is using L'Hospitals Rule?

Could you maybe show me in a little more detail what you mean.

Thank you

6. Originally Posted by Raoh
errr !
$\displaystyle \ln(t)=x\ln(1+\frac{2}{x})$
$\displaystyle \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x}) \right )=2$.
hope it's right now
Could you show me how $\displaystyle \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x}) \right )=2$

Im not really understanding how it equals two

7. Originally Posted by mybrohshi5
Could you show me how $\displaystyle \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x}) \right )=2$

Im not really understanding how it equals two
put $\displaystyle x=\frac{1}{t}$ when $\displaystyle x\to \infty$ we will have $\displaystyle t\to 0^+$.
therefore we should evaluate,
$\displaystyle \lim_{t\to 0^+}$ $\frac{\ln\left(\left 1+2t\right ) \right }{t}$
now use the hospital's rule.
you'll get,
$\lim_{t\to 0^+}\frac{\ln\left (\left 1+2t \right ) \right }{t}=\lim_{t\to 0^+}\frac{2}{1+2t}=2$
hope that right.

8. When i graph the equation it looks like it is approaching about 7.4 when x goes to infinity????

9. Originally Posted by mybrohshi5
When i graph the equation it looks like it is approaching about 7.4 when x goes to infinity????
that's because the limit is $\displaystyle e^2$

10. Could you please show me how to use L'Hospital's Rule to show that it is e^2 cause Naoh tried it and came up with 2

11. He came up with $\displaystyle \lim_{x\to \infty }ln(t)= \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x}) \right )=2$

Now use the fact that $\displaystyle e^{lnx} = x$ to conclude the result.. (remember - you're looking for t)