Tricky limit task

[Ouizzo]

Hello, I have struggled with this task for a while now and been driving stuck. It's mainly how to deal with e^x-1, that I have a problem with. So if some kind soul can help me with this it would be very kind!

Calculate:
lim x->∞ xe^x-1/(x-1)e^x
The main way to do it is to break out the fastest growing term in the numerator and denminator and shorten it.
My solution so far is:

xe^x-1/(x-1)e^x = xe^x-1/xe^x * 1/-e^{x =}

xe^x/e¹/xe^x/1 * 1/-e^x =

1/e¹ * 1/1 * 1/-e^x =

1/e¹*-e<sup>x

I try to paste som nicer equations,</sup> written in word of the task. But it failed once I paste it in???

The answer is: 1/e^x


Thanks in advance!

 

[Ouizzo]

Sorry for the attached picture that dont work out! And the misspelling of denominator!!!

 

[HallsofIvy]

Hello, I have struggled with this task for a while now and been driving stuck. It's mainly how to deal with e^x-1, that I have a problem with. So if some kind soul can help me with this it would be very kind!

Calculate:
lim x->∞ xe^x-1/(x-1)e^x

This is the correct statement of the problem? Then the first thing I would note is that \(\displaystyle \frac{e^{x-1}}{e^x}\) is \(\displaystyle \frac{e^xe^{-1}}{e^x}= e^{-1}\).

So this is exactly \(\displaystyle \lim_{x\to\infty} e^{-1}\frac{x}{x- 1}\).
A good way to handle x, or other variables, going to infinity, it to divide both numerator and denominator by the variable. This limit is exactly \(\displaystyle e^{-1}\lim_{x\to\infty}\frac{\frac{1}{x}}{1- \frac{1}{x}}\)
and, of course, as x goes to infinity, 1/x goes to 0.

The main way to do it is to break out the fastest growing term in the numerator and denminator and shorten it.
My solution so far is:

xe^x-1/(x-1)e^x = xe^x-1/xe^x * 1/-e^{x =}

Frankly, what you are doing here makes no sense. You seem to be saying that \(\displaystyle \frac{xe^{x-1}}{(x- 1)e^x}= \left(\frac{xe^{x-1}}{xe^x}\right)\left(\frac{1}{-e^x}\) and that is simply NOT true. \(\displaystyle \frac{a}{b- c}\) is NOT equal \(\displaystyle \frac{a}{b}\frac{1}{-c}\)

xe^x/e¹/xe^x/1 * 1/-e^x =

1/e¹ * 1/1 * 1/-e^x =

1/e¹*-e<sup>x

I try to paste som nicer equations,</sup> written in word of the task. But it failed once I paste it in???

The answer is: 1/e^x


Thanks in advance!

 

[Deleted member 4993]

Hello, I have struggled with this task for a while now and been driving stuck. It's mainly how to deal with e^x-1, that I have a problem with. So if some kind soul can help me with this it would be very kind!

Calculate:
lim x->∞ xe^x-1)/[(x-1)e^x ]The main way to do it is to break out the fastest growing term in the numerator and denminator and shorten it.
My solution so far is:

xe^x-1/(x-1)e^x = xe^x-1/xe^x * 1/-e^{x =}

xe^x/e¹/xe^x/1 * 1/-e^x =

1/e¹ * 1/1 * 1/-e^x =

1/e¹*-e<sup>x

I try to paste som nicer equations,</sup> written in word of the task. But it failed once I paste it in???

The answer is: 1/e^x ............... Incorrect

Thanks in advance!

\(\displaystyle \displaystyle{\lim_{x \to \infty}\frac{x*e^{x-1}}{(x-1)*e^x}}\)

\(\displaystyle = \displaystyle{\frac{1}{e}*\lim_{x \to \infty}\frac{x*e^{x}}{(x-1)*e^x}}\)

\(\displaystyle = \displaystyle{e^{-1}*\lim_{x \to \infty}\frac{x}{x-1}}\)

\(\displaystyle = \displaystyle{e^{-1}*\lim_{x \to \infty}\frac{1}{1-\frac{1}{x}}}\)

= e^-1

 

[Ouizzo]

Yeah of course. Really great helpful answers guys, thank you so much! your right HallsofIvy I mess up the dividing rules. Im going to take this input with me and analyse them for further similar tasks. And no it was not the right answer I was printed. U did get the right one. Sorry for that!

 

[Ishuda]

Is the problem
\(\displaystyle lim_{x \gt \infty}\dfrac {x e^{x-1}}{(x - 1) e^x}\)?

If that is the case one could write
\(\displaystyle lim_{x \gt \infty}\dfrac {x e^{x-1}}{(x - 1) e^x} = lim_{x \gt \infty}e^{-1}\dfrac {x}{x - 1}\dfrac{e^x}{e^x}
= lim_{x \gt \infty}e^{-1}\dfrac {x}{x - 1} = e^{-1} lim_{x \gt \infty}\dfrac {x}{x - 1} = e^{-1}\).