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$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}+x\right)$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-1-2\sqrt{x^{2}+x}+1+x\right)$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\left(\sqrt{x^{2}+x}+1\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\lim\limits_{x\to\infty}\left(x\sqrt{1+\frac{2}{x}}-1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\lim\limits_{x\to\infty}\left(x\sqrt{1+\frac{1}{x}}+1\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\left(1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\left(\frac{1}{2}\right)+\lim\limits_{x\to\infty}x^{2}= ∞$$

another way I tried:$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}\right)\cdot\frac{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{x\left(-2x^{2}-2x\right)}{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{x\left(-2x^{2}-2x\right)}{x\left(\sqrt{1+\frac{2}{x}}+2\sqrt{1+\frac{1}{x}}\right)}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{\left(-2x^{2}-2x\right)}{\left(\sqrt{1+\frac{2}{x}}+2\sqrt{1+\frac{1}{x}}\right)}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{-2x^{3}-2x}{3}+\lim\limits_{x\to\infty}x^{2}=\lim\limits_{x\to\infty}\frac{x^{3}\left(-2+\frac{3}{x}-\frac{2}{x^{2}}\right)}{3}=-∞$$

and none of them are the answer, why the solutions are not right and can someone use an elementary way to solve the limit?

Absurd
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3 Answers3

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As I mentioned in the comments you cant distribute the limit if it does not exists for instance you cant distribute the limit if the limit of one function is infinity as what you did

The idea to get rid of the negative sign make it positive to do that we need to multiply with the conjugate regroup the positive terms together and the negative together then the following $$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}+x\right)$$

$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}+ x -2\sqrt{x^{2}+x}\right)$$

$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}+x\right)\frac{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}$$ $$\lim_{x\rightarrow \infty} \frac{x\left( x^2 +2x + 2x\sqrt{x^2+2x} + x^2 - 4(x^2+x)\right)}{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}$$ $$\lim_{x\rightarrow \infty} \frac{ x\left( 2x\sqrt{x^2+2x}-2x^2 -2x\right)}{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}$$

$$\lim_{x\rightarrow \infty} \frac{ 2x^2\left( \sqrt{x^2+2x}-(x +1)\right)}{\sqrt{x^{2}+2x}+x+2\sqrt{x^{2}+x}}$$ Multiply with the conjugate again to get $$\lim_{x\rightarrow \infty} \frac{2x^2(x^2+2x - (x+1)^2)}{\left(\sqrt{x^2+2x} + x + 2\sqrt{x^2+x}\right)\left(\sqrt{x^2+2x} + (x+1)\right)} $$ Can you continue ?

IrbidMath
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Here is an elementary way by substitution $x = \frac{1}{t}$ and considering $t\to 0^+$:

\begin{eqnarray*} x\left(\sqrt{x^{2}+2x}- 2\sqrt{x^{2}+x}+x\right) & \stackrel{x=\frac{1}{t}}{=} & \frac{1}{t}\frac{\sqrt{1+2t} - 2\sqrt{1+t} + 1}{t} \\ &= & \left(\frac{\sqrt{1+2t}-1}{t^2} + 2\frac{1-\sqrt{1+t}}{t^2}\right)\\ & = & \frac{2}{t(\sqrt{1+2t}+1)}- \frac{2}{t(1+\sqrt{1+t})} \\ & = & \frac{2}{(1+\sqrt{1+t})((\sqrt{1+2t}+1))}\underbrace{\frac{\sqrt{1+t}-\sqrt{1+2t}}{t}}_{= -\frac{1}{\sqrt{1+t}+\sqrt{1+2t}}}\\ &\stackrel{t \to 0^+}{\longrightarrow} &\frac{2}{4}\cdot \frac{-1}{2} = -\frac{1}{4} \end{eqnarray*}

trancelocation
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Use the binomial theorem for fractional exponents to get $\sqrt{x^2+2x} = x\sqrt{1+\frac{2}{x}}=x\left(1+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{2x^3}-\dots\right)=x+1-\frac{1}{2x}+\frac{1}{2x^2}+\dots).$

Similarly, $\sqrt{x^2+x}=x(1+\frac{1}{x})^{1/2}=x(1+\frac{1}{2x}-\frac{1}{8x^2}+\frac{1}{16x^3}\dots)=x+\frac{1}{2}-\frac{1}{8x}+\frac{1}{16x^2}-\dots.$

Now, using these approximations, the limit can be rewritten as $\lim\limits_{x\to\infty}x((x+1-\frac{1}{2x}+\frac{1}{2x^2}-\dots)-2(x+\frac{1}{2}-\frac{1}{8x}+\frac{1}{16x^2}+\dots)+x),$ which yields a limit of $-\frac{1}{4}.$

Simon Fraser
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