Limits with e^x and an integral too

[G01]

Hi I'm having some trouble with evaluating these limits. I can't figure out what to do. I guess i forgot some calc one. I don't have much work but All I'm asking for for now is a couple hints.

\lim_{x\rightarrow\infty} \frac{e^{3x} -e^{-3x}}{e^{3x} + e^{-3x}}

I tried dividing numerator and denominator by e^3x and 3^-3x Neither worked? Any hints on what else to do?

\lim_{x\rightarrow2^+} e^{3/(2-x)} I have no idea here. Please I know there's not much work but I am totally lost.

Now the integral:

\int e^x \sqrt{1+e^x} dx

I tried the substitution u = \sqrt{1+e^x}

Using that substitution i get this:

2\int u^2 du

integrate this and you get:

(1/2) u^4 + C

which is:

(1+ 2e^x +e^2x)/2, This is nowhere near the answer. Where did I go wrong. I don't see why that substitution didn't work.

 

[G01]

wait i saw my mistake on the integral. Nevermind there.

 

[EbolaPox]

Hello. I'll offer some quick assistance with the integral

\int e^x \sqrt{1+e^x} dx

For this, try setting u = 1 + e^x
Thus,
du = e^x \cdot dx

Does this help?

 

[krab]

try it in a calculator; it converges very quickly. This is because the second term on top and second term on bottom very quickly become utterly negligible.

 

[G01]

Krab and ebola thank you. I solved the integral and the first limit. Now if Anybody can help on the second limit, thatd be great.

 

[Jameson]

\lim_{x\rightarrow2^+} e^{3/(2-x)}

Well as x approaches two, the denominator becomes a very small positive number, let's call it A, and 3/A becomes inifinitely large. Once this infinitely growing number is used as an exponent for e, I would say the limit is positive infinity.

 

[G01]

That not the answer. The answer in the back is 0. I can't figure out how to manipulate that so I can get that answer

 

[benorin]

\lim_{x\rightarrow2^+} e^{3/(2-x)} = e^{\lim_{x\rightarrow2^+} 3/(2-x)} by continuity of the exponential function and \frac{3}{2-x} \rightarrow -\infty \mbox{ as }x\rightarrow2^+ so one may put

\lim_{x\rightarrow2^+} e^{3/(2-x)} = \lim_{u\rightarrow +\infty} e^{u}=0

 

[benorin]

tanh(3x) limit at infinity

\lim_{x\rightarrow\infty} \frac{e^{3x} -e^{-3x}}{e^{3x} + e^{-3x}} =\lim_{x\rightarrow\infty} \frac{e^{3x} -\frac{1}{e^{3x}}}{e^{3x} + \frac{1}{e^{3x}}}=\lim_{x\rightarrow\infty} \frac{\frac{e^{6x}-1}{e^{3x}}}{\frac{e^{6x}+1}{e^{3x}}}=\lim_{x\rightarrow\infty} \frac{e^{6x}-1}{e^{6x}+1}
=\lim_{x\rightarrow\infty} \frac{e^{6x}-1}{e^{6x}+1}\cdot\frac{e^{-6x}}{e^{-6x}}= \lim_{x\rightarrow\infty} \frac{1-e^{-6x}}{1+e^{-6x}}=\frac{1-0}{1+0}=1

 

[VietDao29]

\lim_{x\rightarrow2^+} e^{3/(2-x)} = \lim_{u\rightarrow +\infty} e^{u}=0

Whoops, typo here, benorin.
If u tends to positive infinity, then it should be:
\lim_{u\rightarrow +\infty} e^{u}= + \infty
In this problem, u tends to negative infinity, so it should be:
\lim_{u\rightarrow -\infty} e^{u}= 0 (it's negative infinity, not positive infinity).
By the way, may I suggest you not to post a COMPLETE solution.
If I recalled correctly, this is the third time I catch you posting a COMPLETE solution. :)