Let $\ell =\lim\limits_{x\to b}\frac{f(x)-cb}{g(x)-b^2}$then compute $c-\ell $.

Question

Let $f(x) = xe^2$ and $g(x)= x^{\ln x}$ then prove that the equation

$$f(x)-g(x)=0$$ has two solution $a<b$.

Now for some $c$ we let $$\ell =\lim_{x\to b}\frac{f(x)-cb}{g(x)-b^2}$$

Then what is the value of $c-\ell $?

I was able to show the existence of $a=e^{-1}$ and $b=e^2$. Can someone help me to compute $c-\ell?$

Answer 1

Taking logs, $f(x)=g(x)$ is equivalent to $2+\ln(x)=(\ln(x))^2$, or $t^2=t+2$ where $t=\ln(x)$. It easily follows that $a=e^{-1}$ and $b=e^{2}$.

Notice that the denominator $g(x)-b^2=exp(\ln(x)^2)-e^{4}$ is zero at $x=b$. So for the limit to exist, the numerator must be zero also, otherwise we will have a limit of $-\infty$ on one side and $+\infty$ on the other side.

So $f(b)-cb=0$, whence $c=e^2$. Since $g'(x)=\frac{2\ln(x)}{x}e^{\ln(x)^2}$, we have $g'(b)=4e^2$. By l'Hopital's rule, we then deduce $l=\frac{f'(b)}{g'(b)}=\frac{e^2}{4e^2}=\frac{1}{4}$.

Finally, $c-l$ is $e^2-\frac{1}{4}$.