Thread: Some hard limits I can't solve

1. Some hard limits I can't solve

Hello, I've been solving limits because I have a test in a few days, but a few of them i really cant find... (I study for engineer in Belgium). I expect them to be at least this hard, because they want about 2/3 to fail the 1st year!

Limit x to infiny : ((e^2x) -5x)^(1/x)
Limit x to infinty: x(ln(x+1)-ln(x))

Also, I know limit x to 0: sinx/x is 1, but limit x to zero abs(sinx)/x ?

2. Originally Posted by kev91
Hello, I've been solving limits because I have a test in a few days, but a few of them i really cant find... (I study for engineer in Belgium). I expect them to be at least this hard, because they want about 2/3 to fail the 1st year!

Limit x to infiny : ((e^2x) -5x)^(1/x)
Limit x to infinty: x(ln(x+1)-ln(x))

Also, I know limit x to 0: sinx/x is 1, but limit x to zero abs(sinx)/x ?

The technique for the first is exemplified in this thread: http://www.mathhelpforum.com/math-he...ate-limit.html

For the second, note that $x ( \ln (x + 1) - \ln (x)) = \frac{\ln (x + 1) - \ln (x)}{\frac{1}{x}}$ and now use l'Hopital's Rule.

For the third, the left hand limits and right hand limits are different (-1 and 1) therefore the limit does not exist.

3. Hello, kev91!

The second can be solved with this defintion: . $\lim_{x\to\infty}\left(1+\tfrac{1}{x}\right)^x \;=\;e$

$2)\;\;\lim_{x\to\infty}x\bigg[\ln(x+1)-\ln(x)\bigg]$

We have: . $\lim_{x\to\infty} x\ln\left(\frac{x+1}{x}\right) \;=\;\lim_{x\to\infty}x\left(1 + \tfrac{1}{x}\right) \;=\;\lim_{x\to\infty}\left(1+\tfrac{1}{x}\right)^ x \;=\;e$

4. Originally Posted by Soroban
Hello, kev91!

The second can be solved with this defintion: . $\lim_{x\to\infty}\left(1+\tfrac{1}{x}\right)^x \;=\;e$

We have: . $\lim_{x\to\infty} x {\color{red}\ln} \left(\frac{x+1}{x}\right) \;=\;\lim_{x\to\infty}x\left(1 + \tfrac{1}{x}\right) \;=\;\lim_{x\to\infty}\left(1+\tfrac{1}{x}\right)^ x \;=\;e$
Clearly you're not a lumber jack, Soroban. Because you dropped a (red) log ....

5. Mr fantastic, the 1st limit is to infinity so, I cant use that method you showed me.

Also, I'm prohibitit to use Lhopital for the 2nd limit, but I found it the way Soroban showed me, but a bit differently because he dropped a ln for no reason , the solution is 1. (ln(e))

6. Originally Posted by kev91
Mr fantastic, the 1st limit is to infinity so, I cant use that method you showed me. Mr F says: *Ahem* Yes you can. The indeterminant form is ${\color{red}\frac{\infty}{\infty}}$. See - Wolfram|Alpha and be sure to click on Show steps.

Also, I'm prohibitit to use Lhopital for the 2nd limit, but I found it the way Soroban showed me, but a bit differently because he dropped a ln for no reason , the solution is 1. (ln(e))
Normally I'm not lazy enough to suggest l'Hopital's Rule. But since it is a legitimate method .... (next time, try to mention techniques that you're NOT allowed to use). I suppose you're going to say that you're not allowed to l'Hopital's Rule for the first one either .... (In which case you can note that $\ln (e^{2x} - 5x)$ ~ $2x$ ....)

7. Ok thank you, you have been really helpfull! and yes, I was allowed to use l'hopital in the 1st one .

For this one I have to use Lhopital, but its kind of stupid because I solved it without l'hopital.
lim x to 0 : x^sinx = e ^ lim x to 0: sin x *ln x

sin 0 = 0

and when ln x approaches x=0 is also 0, so the solution is e^0 and that's 1, but it says i HAVE to use l'hopital rule.

Also can you use lhopital when you have 0/ infinity, or does it have to be 0/0 or infinity/infinity.

Sorry for my stupid questions, but they just explained the basics during the class =(, and the test are always way harder!

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