l'Hopital's Rule, part deux

[jtw2e2]

Please tell me where I went wrong:

Find the limit:
lim (xe[sup:dp78fg56]1/x[/sup:dp78fg56] - x)
x->?

So this is going to the form ? - ?. Other than that, I don't even really know how to begin this. The solutions manual advises rewriting as an indeterminate quotient. So I come up with:

lim (e[sup:dp78fg56]1/x[/sup:dp78fg56]-1) / (1/x)
x->?

At this point I tried to use l'Hopital's Rule and find the derivative, but I used the Quotient Rule. I think this is where I went wrong. Please tell me, to use l'Hopital's Rule, I am not actually finding the derivative of the quotient of the functions (as in the Quotient Rule), but simply finding the derivative of the numerator, and the derivative of the denominator??? Forgive me, my professor expects us to acquire this by meditation or psychic powers, neither of which am I adept at.

 

[BigGlenntheHeavy]

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

\(\displaystyle Now, \ the \ final \ limit \ gives \ the \ indeterminate \form \ \frac{0}{0}, \ ergo \ the \ Marqui \ to \ the \ rescue.\)

\(\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here.\)

 

[galactus]

Do you have to use L'Hopital?. If not, here is a way:

\(\displaystyle \lim_{x\to \infty}[xe^{\frac{1}{x}}-x]\)

Let \(\displaystyle h=\frac{1}{x}, \;\ x=\frac{1}{h}\)

Now, making the subs, the limit becomes:

\(\displaystyle \lim_{h\to 0}\frac{e^{h}-1}{h}=1\)

Now, this is the definition of the derivative of e at x=0.

We get 1.


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Here's another way I just scratched out. This is more involved then the above though.

\(\displaystyle \lim_{x\to \infty}[xe^{\frac{1}{x}}-x]\)

Let \(\displaystyle t=\frac{1}{x}, \;\ x=\frac{1}{t}\)

\(\displaystyle \lim_{t\to 0}\frac{e^{t}-1}{t}\)

Let \(\displaystyle u=e^{t}-1\)

\(\displaystyle 1+\frac{1}{u}=e^{t}\)

\(\displaystyle u=ln\left(1+\frac{1}{u}\right)\)

as \(\displaystyle t\to 0\Rightarrow u\to {\infty}\)

Then we get:

\(\displaystyle lim_{u\to \infty}\frac{1}{u\cdot ln(1+\frac{1}{u})}\)

\(\displaystyle =\lim_{u\to \infty}\frac{1}{ln(1+\frac{1}{u})^{u}}\)

Note the famous e limit in the denominator. Then

\(\displaystyle \frac{1}{ln(e)}=1\)

 

[Deleted member 4993]

Do you have to use L'Hopital?. If not, here is a way:


Here's another way I just scratched out. This is more involved then the above though.

\(\displaystyle \lim_{x\to \infty}[xe^{\frac{1}{x}}-x]\)

Let \(\displaystyle t=\frac{1}{x}, \;\ x=\frac{1}{t}\)

\(\displaystyle \lim_{t\to 0}\frac{e^{t}-1}{t}\)

Why not apply L'Hospital right here and get the limit to be 1.

\(\displaystyle \lim_{t\to 0}\frac{e^{t}-1}{t} \ \ = \ \ \lim_{t\to 0}\frac{e^{t}}{1}\ \ = \ \ 1\)

 

[galactus]

I was doing it without L'Hopital. Just a thing of mine.