# Limit as $x$ $\to$ $\infty$ of $x\left(1+\frac1x\right)^x-kx^2\ln\left(1+\frac1x\right)$

Evaluate:- $$\lim_{x\to\infty}\left[ x\left(1+\frac1x\right)^x-kx^2\ln\left(1+\frac1x\right)\right]$$

I tried calculating the limits of the two terms separately. By applying L'Hopital, the 2nd limit can be solved by writing $$x^2\ln\left(1+\frac1x\right)=\frac{\ln\left(1+\frac1x\right)}{\frac{1}{x^2}}$$. Now, L'Hopital gives the final form as $$\frac{x^2}{2(1+x)}=\infty$$. Now, for the 1st term, we know $$\left(1+\frac1x\right)^x$$ is $$e$$ as $$x$$ goes to infinity, so writing the 1st term as $$\frac{\left(1+\frac1x\right)^x}{\frac{1}{x}}$$, we get $$\frac{e}{0}=\infty$$. Now, since both limits are infinite, how can we determine the actual limit?

• The limit = ∞ by graphing on desmos Commented Mar 17 at 5:35
• Take care : the sign of the limit depends on $k$ Commented Mar 17 at 5:44
• Why are you tagging "limits without lhopital"? Commented Mar 17 at 5:58

As $$x\to+\infty$$, $$u:=x\ln\left(1+\frac1x\right)-1\sim-\frac1{2x}\to0$$, and
\begin{align}\left(1+\frac1x\right)^x-kx\ln\left(1+\frac1x\right)&=e^{1+u}-k(1+u)\\&=(e-k)(1+u)+\frac e2u^2+o(u^2)\\ &\begin{cases}\to e-k&\text{if }e\ne k\\\sim\frac e{8x^2}&\text{if }e=k \end{cases} \end{align} hence $$\lim_{x\to\infty}x\left(\left(1+\frac1x\right)^x-kx\ln\left(1+\frac1x\right)\right)=\begin{cases}+\infty&\text{if }e>k\\-\infty&\text{if }e
• How will the limit be $0$ if $e=k$? When splitting the limit,it turns out to be $0 \times \infty$ form,so can we still split the limit? Commented Mar 17 at 8:09
• @a_i_r You just need to evaluate the limit later. For large but finite $x$, the function will be asymptotic to $\frac{e}{8x}$.
• Could you kindly derive the asymptotic form? I actually can't follow you since the big expression(which is multiplied with x) always goes to $e-k$ irrespective of k's value. Commented Mar 17 at 8:53
• It goes to $e-k$ indeed, hence its product by $x$ goes to $\pm\infty$ with the sign of $e-k$ when the latter is $\ne0$, but when $e-k=0$, i.e. when this product is a priori an indeterminate form $\infty\times0$, I solved this indeterminacy by proving that this product is $\sim\frac e{8x}$. What other "asymtptic form" do you want? Commented Mar 17 at 12:02