Why is $\lim\limits_{x \to +\infty}\frac{x \sqrt{x+2}}{\sqrt{x+1}} - x = \frac12$?

Question

I need to evaluate this limit: $$\lim_{x \to +\infty}\frac{x \sqrt{x+2}}{\sqrt{x+1}}-x$$ to calculate the asymptote of this function: $$\frac{x \sqrt{x+2}}{\sqrt{x+1}}$$ which, according to the class notes: $y=x+\frac{1}{2}$ with $a = 1$, $b = \frac{1}{2}$ However, the online math calculators say that currently no steps are available to show for this kind of problem.

I calculated this limit as $x\times \sqrt{1}-x = 1$, but apparently the correct answer is $\frac{1}{2}$.

What is my mistake?

Answer 1

$$ \begin{align}{{x\sqrt{x+2}\over {\sqrt{x+1}}}}-x &=\frac{x\sqrt{x+2}-x\sqrt{x+1}}{\sqrt{x+1}}\\ &=\frac{x\sqrt{x+2}-x\sqrt{x+1}}{\sqrt{x+1}}\left({{{x\sqrt{x+2}+x\sqrt{x+1}}\over {x\sqrt{x+2}+x\sqrt{x+1}}}}\right)\\ &={x\over{\sqrt{x+1}(\sqrt{x+2}+\sqrt{x+1})}}\\ &={1\over{\sqrt{1/x+1}(\sqrt{1+2/x}+\sqrt{1/x+1})}} \end{align}$$

so the limit at $+\infty$ is $1/2$.

Answer 2

Note that $$\frac{\sqrt{x+2}}{\sqrt{x+1}}=\sqrt{1+\frac1{x+1}}\approx\sqrt{1+2\cdot\frac1{2(x+1)}+\left(\frac1{2(x+1)}\right)^2}=1+\frac1{2(x+1)} $$ (of course you have to be more explicit about the "$\approx$" in your work)

Answer 3

I assume the limit is for $x\to\infty$. (And yes, you should specify that sort of things... $x\to0$ would equally make sense!)

Are you familiar with Taylor expansions? If so,$^{(\dagger)}$ essentially what you want is immediately implied by the following: for $x>0$, $$\begin{align} \frac{\sqrt{x+2}}{\sqrt{x+1}} &= \frac{\sqrt{1+\frac{2}{x}}}{\sqrt{1+\frac{1}{x}}} = \frac{1+\frac{2}{2x}+o\left(\frac{1}{x}\right)}{1+\frac{1}{2x}+o\left(\frac{1}{x}\right)} = ({1+\frac{1}{x}+o\left(\frac{1}{x}\right)})({1-\frac{1}{2x}+o\left(\frac{1}{x}\right)})\\ &= {1+\frac{1}{2x}+o\left(\frac{1}{x}\right)} \end{align}$$ when $x\to\infty$, using Taylor expansions at $0$ (since $\frac{1}{x}\xrightarrow[x\to\infty]{} 0$).

It follows that $$ x\left(\frac{\sqrt{x+2}}{\sqrt{x+1}} - 1\right) = x\left(\frac{1}{2x}+o\left(\frac{1}{x}\right)\right) = \frac{1}{2}+o\left(1\right)\xrightarrow[x\to\infty]{} \frac{1}{2} $$

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$(\dagger)$ If not, I strongly encourage you to look into these when you feel you have enough background. They form a powerful, versatile, systematic tool to compute limits, asymptotics, and estimates. Pretty much a Swiss army knife, only with Landau notations instead of a corkscrew.

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Adding more on your specific mistake. You have $x\cdot f(x) - x$, where $f(x) \to 1$ (here, I write $f(x) = \frac{\sqrt{x+2}}{\sqrt{x+1}}$ for convenience). You somehow "calculate" this as $x\cdot \sqrt{1}-x=1$ (how?), but this is incorrect. More precisely, you can rewrite $$xf(x)-x = x(f(x)-1) "\to" \infty\cdot 0$$so you have an indeterminate form... and you need to get into more details about how $f(x)$ approaches its limit when $x\to \infty$ to conclude. (It turns out that when $x$ is big, $f(x) - 1 \simeq \frac{1}{2x}$, hence the limit...)

Answer 4

Here's a simple approach. First of all sum the two fractions: $$L := \lim_{x \to +\infty} \frac{x\sqrt{x + 2}}{\sqrt{x+1}} - x = \lim_{x \to +\infty} \frac{x\sqrt{x + 2} - x\sqrt{x + 1}}{\sqrt{x + 1}}$$

With square roots it's usually the best to multiply and divide by the conjugate (in order to avoid indeterminate forms altogether): $$L = \lim_{x \to +\infty} \frac{x(\sqrt{x + 2} - \sqrt{x + 1})}{\sqrt{x+1}} = \lim_{x \to +\infty} x\frac{\sqrt{x + 2} - \sqrt{x + 1}}{\sqrt{x + 1}}\frac{\sqrt{x + 2} + \sqrt{x + 1}}{\sqrt{x + 2} + \sqrt{x + 1}}$$ Then: $$\require{cancel}L = \lim_{x \to +\infty} \frac{\bcancel{\color{red}x}(\cancel{\color{green}x} + 2 - \cancel{\color{green}x} - 1)}{\bcancel{\color{red}x}\underbrace{\sqrt{1 + \frac1x}}_{\to 1}\underbrace{\left(\sqrt{1 + \frac2x} + \sqrt{1 + \frac1x}\right)}_{\to 2}} = \frac12$$