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

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