Wrong result for a $1^\infty$ limit

Question

I wanna know the $$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x .$$ I applied $1^\infty$ algorithm and I have in last sentence $$\lim_{x\rightarrow\infty} \left(\frac{2x+4-\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right)^x.$$ After that I used $x=e^{\ln x}$ and got this: $$\lim_{x\rightarrow\infty} \left(\frac{x^2+4x-x\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right).$$ The result for this, I think is $2$, but I'm not really sure of this, Wolfram-Alpha says that it's $\frac{3}{2}$. Can explain somebody why, but without L'Hospital?

Thks a lot!

Answer 1

Sometimes WA gives absolutely wrong results, but in our case he gave a right answer.

Since $f(x)=e^x$ is a continuous function, we obtain: $$\lim_{x\rightarrow+\infty}(\sqrt{x^2+2x+4}-x)^x =\lim_{x\rightarrow+\infty}\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}\right)^x =$$ $$=\lim_{x\rightarrow+\infty}\left(1+\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)^{\frac{1}{\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1}\cdot x\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)} =$$ $$=e^{\lim\limits_{x\rightarrow+\infty}x\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)}=e^{\lim\limits_{x\rightarrow+\infty}\frac{x( x+4-\sqrt{x^2+2x+4})}{\sqrt{x^2+2x+4}+x}}=$$ $$=e^{\lim\limits_{x\rightarrow+\infty}\frac{x(6x+12)}{\left(\sqrt{x^2+2x+4}+x\right)\left( x+4+\sqrt{x^2+2x+4}\right)}}=e^{\frac{6}{2\cdot2}}=e^{\frac{3}{2}}.$$

Answer 2

This is evaluated with Taylor's formula at order $2$: $$\ln\Bigl(\bigl(\sqrt{x^2+2x+4}-x\bigr)^x\Bigr)=x\ln\bigl(\sqrt{x^2+2x+4}-x\bigr)=x\Bigl(\ln x+\ln\biggl(\sqrt{1+\frac2x+\frac4{x^2}}-1\biggr)\biggr). $$ Now, Taylor's expansion at order $2$ yields \begin{align}\sqrt{1+\frac2x+\frac4{x^2}}-1&=1+\frac12\biggl(\frac2x+\frac4{x^2}\biggr)-\frac18\biggl(\frac2x+\frac4{x^2}\biggr)^2+o\biggl(\frac1{x^2}\biggr)-1\\ &=\frac1x+\frac3{2x^2}+o\biggl(\frac1{x^2}\biggr)=\frac1x\biggl(1+\frac3{2x}+o\biggl(\frac1{x}\biggr)\biggr),\end{align} so that $$\ln\biggl(\sqrt{1+\frac2x+\frac4{x^2}}-1\biggr)=-\ln x+\ln\biggl(1+\frac3{2x}+o\biggl(\frac1{x}\biggr)\biggr)=-\ln x+\frac3{2x}+o\biggl(\frac1{x}\biggr),$$ and finally $$\ln\Bigl(\bigl(\sqrt{x^2+2x+4}-x\bigr)^x\Bigr)=x\biggl(\ln x-\ln x+\frac3{2x}+o\biggl(\frac1{x}\biggr)\biggr)=\frac32+o(1)\to\frac32.$$

Answer 3

Want $\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x $.

Since, for small $z$, $\sqrt{1+z} =1+\frac12 z + O(z^2) $ and $\ln(1+z) =z+O(z^2) $,

$\begin{array}\\ \sqrt{x^2+2x+4} &=\sqrt{(x+1)^2+3}\\ &=(x+1)\sqrt{1+\frac{3}{(x+1)^2}}\\ &=(x+1)(1+\frac{3}{2(x+1)^2}+O(x^{-4}))\\ &=x+1+\frac{3}{2(x+1)}+O(x^{-3})\\ \text{so}\\ \sqrt{x^2+2x+4}-x &=1+\frac{3}{2(x+1)}+O(x^{-3})\\ \end{array} $

Therefore $\ln(\sqrt{x^2+2x+4}-x) =\ln(1+\frac{3}{2(x+1)}+O(x^{-3})) =\frac{3}{2(x+1)}+O(x^{-2}) $ so

$\begin{array}\\ x\ln(\sqrt{x^2+2x+4}-x) &=\frac{3x}{2(x+1)}+O(x^{-1})\\ &=\frac{3(x+1)-3}{2(x+1)}+O(x^{-1})\\ &=\frac32-\frac{3}{2(x+1)}+O(x^{-1})\\ &=\frac32+O(x^{-1})\\ \end{array} $

so that $\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x =e^{3/2} $.