# Thread: find limits using L'Hopital rule and FTC

1. ## find limits using L'Hopital rule and FTC

Please kindly check if there are any mistakes.

1. Find $\displaystyle \lim_{x\rightarrow+\infty}e^{-e^x}\int_0^x e^{e^t} dt$

Solution.
I use Fundamental Theorem of Calculus (FTC) $\displaystyle \frac{d}{dx}\int_0^x f(t)dt=f(x)-f(0)$ to simplify the integral and then L'Hopital rule to find the limit:

$\displaystyle \lim_{x\rightarrow\infty}e^{-e^x}\int_0^x e^{e^t}dt=\lim_{x\rightarrow\infty}\frac{e^{e^x}-e^{e^0}}{e^{e^x}e^x}=\lim_{x\rightarrow\infty}\fra c {e^{e^x}-e^{e^0}}{e^{e^x}e^x}=$

$\displaystyle =\lim_{x\rightarrow\infty}\frac{e^{e^x}-e}{e^{e^x}e^x}=\lim_{x\rightarrow\infty}\frac{e^{e ^x}(1-\frac{e}{e^{e^x}})}{e^{e^x}e^x}=\lim_{x\rightarrow \infty}\frac{1-e/\infty}{e^x}=\lim_{x\rightarrow\infty}\frac{1-0}{\infty}=0$

2. Use L'Hopital's Rule to find the following limits:

(i) $\displaystyle lim_{x\rightarrow0}xlnx$

(ii) $\displaystyle lim_{x\rightarrow0}x^x$

(iii) $\displaystyle lim_{x\rightarrow0}\frac{1}{x}\int_0^x t^t dt$

Deduce the value of the limit $\displaystyle lim_{x\rightarrow0}ln x \int_0^x t^t dt$.

Solution.

(i) substitute $\displaystyle y=\frac{1}{x}$ and take first derivative

$\displaystyle lim_{x\rightarrow0}xlnx=lim_{y\rightarrow\infty}\f rac{ln1-lny}{y}=lim_{y\rightarrow\infty}\frac{(-ln y)}{y}=lim_{y\rightarrow\infty}(-\frac{1}{y})=0$

(ii) $\displaystyle lim_{x\rightarrow0}x^x$
Take logarithm of the [tex]x^x{/math] and its limit:
$\displaystyle \ln(x^x)=xlnx$ and $\displaystyle \lim_{x\rightarrow0} xlnx=0$, see (i).
Therefore if $\displaystyle lim_{x\rightarrow0}\ln(x^x)=0$, then the expression under log sign should be $\displaystyle \rightarrow1$.

does this comment make sense?...

(iii) $\displaystyle \lim_{x\rightarrow0}\frac{1}{x}\int_0^x t^t dt$

Using L'Hopital Rule and FTC,

$\displaystyle =\lim_{x\rightarrow0}\frac{\frac{d}{dt}\int_0^xt^t dt}{x}=\lim_{x\rightarrow0}(x^x-0^0)=1-1=0$

Here I am not sure at all whether I can take 0^0 - I rely on (ii) where I proved that $\displaystyle lim_{x\rightarrow0}x^x=1$.

Therefore, $\displaystyle lim_{x\rightarrow0}ln x \int_0^x t^t dt$=\

$\displaystyle lim_{x\rightarrow0}lnx*\lim_{x\rightarrow0}\int_0^ x t^t dt=1*0=0$

2. Originally Posted by Volga
$\displaystyle \frac{d}{dx}\int_0^x f(t)dt=f(x)-f(0)$
It should be:

$\displaystyle \dfrac{d}{dx}\displaystyle\int_0^x f(t)\;dt=f(x)$

Fernando Revilla

3. Originally Posted by Volga
$\displaystyle lim_{x\rightarrow0}xlnx=lim_{y\rightarrow\infty}\f rac{ln1-lny}{y}=lim_{y\rightarrow\infty}\frac{(-ln y)}{y}=lim_{y\rightarrow\infty}(-\frac{1}{y})=0$
Right.

Therefore if $\displaystyle lim_{x\rightarrow0}\ln(x^x)=0$, then the expression under log sign should be $\displaystyle \rightarrow1$.
does this comment make sense?...
Better, suppose:

$\displaystyle L=\displaystyle\lim_{x \to a}{f(x)}\;,\quad\lambda=\displaystyle\lim_{x \to a}{\log f(x)}$

the, using the continuity of $\displaystyle \log$:

$\displaystyle e^{\lambda}=e^{\displaystyle\lim_{x \to a}{\log f(x)}}=e^{\log \left({\displaystyle\lim_{x \to a}{ f(x)}\right)}=\displaystyle\lim_{x \to a}{f(x)}=L$

$\displaystyle =\lim_{x\rightarrow0}\frac{\frac{d}{dt}\int_0^xt^t dt}{x}=\lim_{x\rightarrow0}(x^x-0^0)=1-1=0$
See my previous post. Besides, $\displaystyle 0^0$ is an indetermined expression.

Fernando Revilla

4. Originally Posted by FernandoRevilla
It should be:

$\displaystyle \dfrac{d}{dx}\displaystyle\int_0^x f(t)\;dt=f(x)$

Fernando Revilla
I am glad it is! This was my first version (I think I took it from one of the books) then I started having doubts and added f(0) by analogy with f(b)-f(a) when the limits of integration are a and b. I guess I need to work it out myself to stop having doubts.

For $\displaystyle 0^0$, I googled for 0^0 and (1) google answered "1" and (2) there was this article which said "As a rule of thumb, one can say that 0^0 = 1".

sci.math FAQ: What is 0^0?

Conclusion: when it comes to mathematics, google is not a good source )))

5. Oh, is that because the derivative of a constant is zero?...

So d/dt of F(0) is 0, no matter what function F is

6. Originally Posted by Volga
For $\displaystyle 0^0$, I googled for 0^0 and (1) google answered "1" and (2) there was this article which said "As a rule of thumb, one can say that 0^0 = 1".
When the expression $\displaystyle 0^0$ "comes from trying to find a limit" you can't say $\displaystyle 0^0=1$. You need additional information about the corresponding functions.

Originally Posted by Volga
Oh, is that because the derivative of a constant is zero?... So d/dt of F(0) is 0, no matter what function F is

Please, look at the proof of the Fundamental Theorem of Calculus.

Fernando Revilla

7. Just as a side note: there are way too many questions in the OP for one thread. See Rule # 8.

8. Looked up the proof, no more questions. Thank you for your instruction!

I will try not to keep to the limit of # of questions in a thread in the future. These ones all seemed to be related...