Question

http://tinypic.com/r/2zq6kd3/6

Answer 1

I've gotten as far as the integration, but when I try to do the limit to infinity I get stuck

Answer 2

Okay I'm really stuck here. My integration came out as 4/sqrt3 * ln [ (sqrt(3+x^2))/x - sqrt(3)/x ] from 1 to t->infinity

Answer 3

oh and change the 3 at the top of the original problem to a 4

Answer 4

\[\frac{4}{\sqrt{3}}\ln\left(\frac{\sqrt{3+x^2}}{x} - \frac{\sqrt{3}}{x}\right)\]

From 1 to t->infinity:
\[\lim_{t\rightarrow\infty}\left(\frac{4}{\sqrt{3}}\ln\left(\frac{\sqrt{3+t^2}}{t} - \frac{\sqrt{3}}{t}\right) - \frac{4}{\sqrt{3}}\ln\left(\frac{\sqrt{3+1^2}}{1} - \frac{\sqrt{3}}{1}\right)\right)\]

\[\frac{4}{\sqrt 3}\lim_{t\rightarrow\infty}\ln\left(\frac{\sqrt{3+t^2}}{t} - \frac{\sqrt{3}}{t}\right) -\frac{4}{\sqrt 3}\ln\left(\frac{\sqrt{3+1^2}}{1} - \frac{\sqrt{3}}{1}\right)\]

\[\frac{4\sqrt3}{3}\ln\left( \lim_{t\rightarrow\infty}\left( \frac{\sqrt{3+t^2}}{t} - \frac{\sqrt{3}}{t} \right) \right) - \frac{4}{\sqrt 3}\ln\left(\frac{2-\sqrt{3}}{1}\right)\]

Can you go on from here?

Answer 5

so you can just move things around?
would we use the common denominator?
technically sqrt3 - sqrt3 is zero, and t/t is 1 right?

Answer 6

The indefinite integral is (according to wolfram),
\[\frac{4}{\sqrt 3}\left(\ln{x}-\ln\left(\sqrt3\sqrt{x^2+3}+3\right)\right)\]
your solution looks different, but that might just be different simplifications made. I'll use the wolfram one just to show the steps since you asked about how to evaluate the limit, and the steps should be pretty much the same :)

So the definite integral from 1 to t is

\[\frac{4}{\sqrt 3}\left(\ln{\frac{t}{\sqrt3\sqrt{t^2+3}+3}}\right)-\frac{4}{\sqrt 3}\left(\ln{\frac{1}{2\sqrt3+3}}\right)\]

which gives:
\[\lim_{t\rightarrow\infty}\left(\frac{4}{\sqrt 3}\left(\ln{\frac{t}{\sqrt3\sqrt{t^2+3}+3}}\right)-\frac{4}{\sqrt 3}\left(\ln{\frac{1}{2\sqrt3+3}}\right)\right)\]

\[\lim_{t\rightarrow\infty}\left(\frac{4}{\sqrt 3}\left(\ln{\frac{t}{\sqrt3\sqrt{t^2+3}+3}}\right)\right)-\lim_{t\rightarrow\infty}\frac{4}{\sqrt 3}\left(\ln{\frac{1}{2\sqrt3+3}}\right)\]

Since the second limit is just a constant this becomes:
\[\lim_{t\rightarrow\infty} \left(\frac{4}{\sqrt 3}\left(\ln{\frac{t}{\sqrt3\sqrt{t^2+3}+3}}\right)\right)-\frac{4}{\sqrt 3}\left(\ln{\frac{1}{2\sqrt3+3}}\right)\]
\[\frac{4}{\sqrt 3}\lim_{t\rightarrow\infty}\ln{\frac{t}{\sqrt3\sqrt{t^2+3}+3}}-\frac{4}{\sqrt 3}\left(\ln{\frac{1}{2\sqrt3+3}}\right)\]

Now if \[\lim_{t\rightarrow\infty}\frac{t}{\sqrt3\sqrt{t^2+3}+3}=L\]
exists and ln(t) is continuous at L we can write it as:
\[\frac{4}{\sqrt 3}\ln{\lim_{t\rightarrow\infty} \left(\frac{t}{\sqrt3\sqrt{t^2+3}+3}\right)}-\frac{4}{\sqrt 3}\left(\ln{\frac{1}{2\sqrt3+3}}\right)\]

So we start by evaluating
\[\lim_{t\rightarrow\infty}\frac{t}{\sqrt3\sqrt{t^2+3}+3}\]
Using L'Hospital's rule this becomes:
\[\lim_{t\rightarrow\infty}\frac{\sqrt{t^2+3}}{\sqrt3 t}\]
Factoring out constants and simplifying gives
\[\frac{1}{\sqrt{3}}\lim_{t\rightarrow\infty} \sqrt{\frac{t^2 + 3}{t^2}}\]
Using power rule and L'Hospital's rule again (since it's still infinity/infinity) we get
\[\frac{1}{\sqrt 3}\sqrt{\lim_{t\rightarrow\infty} \frac{2t}{2t}} = \frac{1}{\sqrt3 }\]

Since ln(t) is continuous at t = 1/sqrt(3) we can write the original limit as
\[\frac{4}{\sqrt 3}\ln{\lim_{t\rightarrow\infty} \left(\frac{t}{\sqrt3\sqrt{t^2+3}+3}\right)}-\frac{4}{\sqrt 3}\ln{\frac{1}{2\sqrt3+3}}\]
We already have the limit so we just plug that in to get the answer
\[\frac{4}{\sqrt 3}\ln{\frac{1}{\sqrt3 } }-\frac{4}{\sqrt 3}\ln{\frac{1}{2\sqrt3+3}}\]

Answer 7

Your integral is also correct since wolfram alpha gives the same limit, and the steps should be similar :)