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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?

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    $\begingroup$ See this. That may help to make your question more readable. As it stands right now, it is a mess. $\endgroup$ Commented Jul 28, 2016 at 20:15
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    $\begingroup$ You should start with stating which limit you are trying to find. $x\to\infty$, $x\to 0$, $x\to x_0$, or something else entirely. Otherwise it difficult to understand what you are talking about. $\endgroup$ Commented Jul 28, 2016 at 20:21
  • $\begingroup$ By the way, the answers currently fail to point out your mistake: you have $x\times f(x) - x$, where $f(x) \to 1$. You somehow "calculate" this as $x\times \sqrt{1}-x=1$ (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...) [Added to my answer, as comments are not meant to last.] $\endgroup$
    – Clement C.
    Commented Jul 28, 2016 at 21:04
  • $\begingroup$ hi Thx for all the improvements in the question and the answers. I did not expect someone to answer actually, so I didn't spent a whole lot of time posing the question. My bad. As the question is the calculate a sledged straight asymptote (y=ax+b) to f(x) I guess the limit is for x to + or - infinity indeed. $\endgroup$
    – babipsylon
    Commented Jul 29, 2016 at 15:04

4 Answers 4

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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: $$\require{cancel}L = \lim_{x \to +\infty} \frac{\bcancel{\color{red}x}(\cancel{\color{green}x} + 2 - \cancel{\color{green}x} - 1)}{\bcancel{\color{red}x}\underbrace{\sqrt{1 + \frac1x}}_{\to 1}\underbrace{\left(\sqrt{1 + \frac2x} + \sqrt{1 + \frac1x}\right)}_{\to 2}} = \frac12$$

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

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

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  • $\begingroup$ I took the liberty to improve the formatting of your answer. Let me know if this looks fine to you. $\endgroup$
    – Clement C.
    Commented Jul 28, 2016 at 20:59
  • $\begingroup$ it is ok with me $\endgroup$ Commented Jul 28, 2016 at 21:00
  • $\begingroup$ I don't understand the last step. It seems you divided both num and denum by x, but 1/x under the square root, it should become 1/x^2, no? Also, when doing (a*b)/c, the c should only applied to a or b, not to both a and b. Here a = sqrt(x+1) and b = sqrt(x+2)+sqrt(x+1). $\endgroup$
    – babipsylon
    Commented Jul 29, 2016 at 19:54
  • $\begingroup$ @TsemoAristide Thx. The only thing I still don't get from this is how to derive that sqrt(x+1) = sqrt(x*(x+(1/x))). It seems to me that it should be sqrt(x*(1+(1/x)))? You did it like that at the right side , but not at the left side of the denominator. But I guess it's a typo, because in your final solution above it is correct. $\endgroup$
    – babipsylon
    Commented Jul 29, 2016 at 20:19
  • $\begingroup$ Yes it is a typo $\endgroup$ Commented Jul 29, 2016 at 20:20
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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 $$ x\left(\frac{\sqrt{x+2}}{\sqrt{x+1}} - 1\right) = x\left(\frac{1}{2x}+o\left(\frac{1}{x}\right)\right) = \frac{1}{2}+o\left(1\right)\xrightarrow[x\to\infty]{} \frac{1}{2} $$


$(\dagger)$ 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.


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...)

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  • $\begingroup$ Hi, thanks, but I don't need to know the Taylor expansion thing for the classes for now. $\endgroup$
    – babipsylon
    Commented Jul 29, 2016 at 20:06
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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)

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