MHB Limit problem with exponents/logarithms

[GreenGoblin]

Hello,

I am tackling a problem which I have an answer for but I can't complete the work of showing where it's coming from. This is where I have arrived

$\lim \frac{e^{xln(3+x)}}{e^{xlnx}}$ for x goes to infinity. (Also, what's the code for this please? To show the lim with arrow?)

I have that the answer is $e^{3}$ but not how to get there. I've tried a lot of cancelling but not quite got there.

Gracias,
GreenGoblin

 

[alexmahone]

Hello,

I am tackling a problem which I have an answer for but I can't complete the work of showing where it's coming from. This is where I have arrived

$\lim \frac{e^{xln(3+x)}}{e^{xlnx}}$ for x goes to infinity. (Also, what's the code for this please? To show the lim with arrow?)

I have that the answer is $e^{3}$ but not how to get there. I've tried a lot of cancelling but not quite got there.

Gracias,
GreenGoblin

$\displaystyle\lim_{x\to\infty}\frac{e^{x\ln(3+x)}}{e^{x\ln x}}=\lim_{x\to\infty}\frac{e^{\ln(3+x)^x}}{e^\ln x^x}$

$\displaystyle=\lim_{x\to\infty}\frac{(3+x)^x}{x^x}$

$\displaystyle=\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^x$

$\displaystyle=\lim_{x/3\to\infty}\left[\left(1+\frac{3}{x}\right)^{x/3}\right]^3$

$\displaystyle=e^3$

 

[GreenGoblin]

Hi,

Your third line is actually the start of the problem, I had arrived at what I wrote by rearranging from that. Are you telling me I was moving away from a solution? Now I just don't see at all where $e^{3}$ comes from as a solution from that.

 

[alexmahone]

Hi,

Your third line is actually the start of the problem, I had arrived at what I wrote by rearranging from that. Are you telling me I was moving away from a solution? Now I just don't see at all where $e^{3}$ comes from as a solution from that.

Do you know that $\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$?

 

[GreenGoblin]

Yes but I don't think such standard results are allowed to be used in this case. This problem comes from a set of exercises on L'Hospital's rule, hence why I looked to get in a fraction form (even if it is not applicable to use this rule here).

 

[polygamma]

$ \displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x} \right)^{x} $

let $ \displaystyle y = \ln \left(1 + \frac{3}{x} \right)^{x} = x \ln \left(1 + \frac{3}{x} \right) = \frac{\ln \left(1 + \frac{3}{x} \right)}{\frac{1}{x}} $

$\displaystyle \lim_{x \to \infty} y = \lim_{x \to \infty} \frac{\ln \left(1 + \frac{3}{x} \right)}{\frac{1}{x}} = \lim_{x \to \infty} \frac{\frac{1}{1+\frac{3}{x}} \left(\frac{-3}{x^{2}}\right)}{-\frac{1}{x^{2}}} = \lim_{x \to \infty} \frac{3}{1+\frac{3}{x}} =3 $

therefore $\displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x} \right)^{x}= \lim_{x \to \infty} e^{y} = e^{\lim x \to \infty y} = e^{3}$