Calculate $\;\lim\limits_{x\to\infty} \left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]$.
It is a $\frac00$ case of indetermination if we rewrite as $\lim_{x\to\infty} \frac{((1+\frac1x)^x-e\ln(1+\frac1x)^x)}{\frac{1}{x^2}}$, since $\lim_{x\to\infty} \frac{1}{x^2} = 0$, $\lim_{x\to\infty} (1+\frac1x)^x = e$ and $\lim_{x\to\infty} \ln(1+\frac1x)^x = 1$.
I think that it is the type that has a solution without l'Hospital's rule, but it's quite difficult to find, so l'Hospital still remains the best try to me. I tried using it with different rewrites, but it seems that it needs to be used multiple times, and the expression gets harder and harder to calculate, so I assume that some other limit must be applied first to make the expression nicer.
Also, I futilely tried to use the following known limits by changing $x$ into $y = \frac1x$ if needed (and adding and substracting $ex^2$ in the main parenthesis and trying to use the last 2 limits), but maybe it can help you: $\lim_{x\to0} \frac{a^x-1}{x} = \ln a$, $\lim_{x\to0} \frac{\ln(1+x)}{x} = 1$, $\lim_{x\to0} \frac{(1+x)^r-1}{x} = r$, $\lim_{x\to0} \frac{(1+x)^\frac1x-e}{x} = -\frac{e}{2}$, $\lim_{x\to\infty} (x-x^2ln(1+\frac1x)) = \frac12$.
Can you help me with this problem?
Best Answer
I am going to calculate your limit without using Taylor series but only the two following notable limits:
$\lim\limits_{x\to0}\dfrac{\ln(1+x)}{x}=1\;,\quad\lim\limits_{x\to0}\dfrac{x-\ln(1+x)}{x^2}=\dfrac12\;.$
$\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]=$
$=\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^2\ln\left(1+\dfrac1x\right)^x\right]=$
$=\lim\limits_{x\to\infty}\dfrac{\left(1+\frac1x\right)^x-e\ln\left(1+\frac1x\right)^x}{\left[1-\ln\left(1+\frac1x\right)^x\right]^2}\cdot\lim\limits_{x\to\infty}\left[x-x\ln\left(1+\frac1x\right)^x\right]^2$
Now I am going to calculate the first limit by using the following substitution:
$t=\dfrac1e\left(1+\dfrac1x\right)^x-1\;.$
$\lim\limits_{x\to\infty}\dfrac{\left(1+\dfrac1x\right)^x-e\ln\left(1+\dfrac1x\right)^x}{\left[1-\ln\left(1+\dfrac1x\right)^x\right]^2}=$
$=\lim\limits_{t\to0}\dfrac{e(1+t)-e\ln\big[e(1+t)\big] }{\left\{1-\ln\big[e(1+t)\big]\right\}^2}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{1+t-1-\ln(1+t)}{\big[1-1-\ln(1+t)\big]^2}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{t-\ln(1+t)}{\ln^2(1+t)}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{t-\ln(1+t)}{t^2}\cdot\lim\limits_{t\to0}\left[\dfrac{t}{\ln(1+t)}\right]^2=$
$=e\cdot\dfrac12\cdot1^2=\dfrac e2\;.$
Now I am going to calculate the second limit by using the following substitution:
$u=\dfrac1x\;.$
$\lim\limits_{x\to\infty}\left[x-x\ln\left(1+\dfrac1x\right)^x\right]^2=$
$=\lim\limits_{x\to\infty}\left[x-x^2\ln\left(1+\dfrac1x\right)\right]^2=$
$=\lim\limits_{u\to0}\left[\dfrac1u-\dfrac1{u^2}\ln\big(1+u\big)\right]^2=$
$=\lim\limits_{u\to0}\left[\dfrac{u-\ln(1+u)}{u^2}\right]^2=$
$=\left(\dfrac12\right)^2=\dfrac14\;.$
Consequently,
$\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]=$
$=\dfrac e2\cdot\dfrac14=\dfrac e8\;.$