Find $\lim_{x\to\infty} (x-\sqrt{x^2+x+1}\frac{\ln(e^x+x)}{x})$.

Question

Find $\lim_{x\to\infty} (x-\sqrt{x^2+x+1}\frac{\ln(e^x+x)}{x})$.

I did $\lim_{x\to\infty} (x-\sqrt{x^2+x+1}\frac{\ln(e^x+x)}{x})$=$\lim_{x\to\infty} (\frac{x^2-\sqrt{x^2+x+1}\ln(e^x+x)}{x})$

$(e^x+x)^\frac{1}{x}=e(1+x/e^x)^\frac{1}{x}$. Now making $y=x/e^x$, we obtain $e(1+y)^\frac{e^{-x}}{y}=e((1+y)^\frac{1}{y})^{e^{-x}}$ and as $y(x)\rightarrow{0}$, $\lim_{x\to\infty}\frac{\ln(e^x+x)}{x}=e$ and I think the limit is $-\frac{1}{2}$. However, this is just a quess or a sketch of a proof. Any help please?

Answer 1

Let $L$ be the limit. We split it into two parts $L=L_1+L_2$ with

$$ L_1=\lim_{x\to\infty}\left(x-\ln(\mathrm e^x+x)\right),\qquad L_2=\lim_{x\to\infty}\left(\ln(\mathrm e^x+x)-\sqrt{x^2+x+1}\cdot\frac{\ln(\mathrm e^x+x)}x\right) $$

Then

$$ L_1=\lim_{x\to\infty}\left(x-\ln(\mathrm e^x)-\ln\left(1+\frac x{\mathrm e^x}\right)\right)=-\lim_{x\to\infty}\ln\left(1+\frac x{\mathrm e^x}\right)=0 $$

and

$$ L_2=\lim_{x\to\infty}\underbrace{\frac{\ln(\mathrm e^x+x)}x}_{\to 1}\cdot\underbrace{\left(x-\sqrt{x^2+x+1}\right)}_{\to-\frac 12}=-\frac 12. $$

Hence $L=-\frac 12$.

Answer 2

Let $\phi(x)\,=\,1+\frac{1}{x}+\frac{1}{x^2}.$ Note $\lim\phi(x)\,=\,\lim\sqrt{\phi(x)}\,=\,1,$ and $$\left(\sqrt{\phi(x)}\right)'\,=\,\frac{-\sqrt{\phi(x)}}{2\phi(x)}\left(\frac{1}{x^2}\,+\,\frac{2}{x^3}\right).$$

Let $z(x)\,=\,\ln(1+x e^{-x})$, so that $\lim z(x)\,=0.$ Let $$p(x)\,=\,x-x\sqrt{\phi(x)}\,=\,\frac{1-\sqrt{\phi(x)}}{\frac{1}{x}},$$ so that $p(x)$ has form "$\frac{0}{0}$" as $x \to +\infty$. By L'hopital rule, we get $$\lim p(x)\,=\,\lim\frac{-\sqrt{\phi(x)}}{2\phi(x)}\left(1\,+\,\frac{2}{x}\right)\,=\,\frac{-1}{2}.$$

Now your expression is $$x-\sqrt{\phi(x)}\ln(e^x+x)\,=\,x-\sqrt{\phi(x)}\ln\left( e^x(1+xe^{-x}) \right)$$

$$ =\,x-\sqrt{\phi(x)}\left(x\,\,+\,\,\ln(1+xe^{-x}) \right)\,=\,\,p(x)-z(x)\sqrt{\phi(x)}\,\to\,\frac{-1}{2}.$$