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I'm trying to find oblique asmyptotes for the function $\sqrt{x^2+x+1}$ and I manage to caclculate that the coefficient for the asymptote when x approaches infinity is 1.

But when i try to find the m-value for my oblique asymptote by taking the limit of:

$$ \lim_{x\to\infty}\sqrt{x^2+x+1}-x=m $$

I'm stuck.

How do i simplify this expression to find the limit?

I've tried manipulating the radical by converting it to the denominator:

$$\lim_{x\to\infty}\frac{x^2+x+1}{\sqrt{x^2+x+1}}-x.$$

Or by multiplying both terms with the conjugate:

$$\lim_{x\to\infty}\frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}}$$

But in neither case do I know how to take the limit. Any help would be greatly appreciated.

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  • $\begingroup$ $\lim_{x\rightarrow\infty}\sqrt{x^2+x+1} \neq 1$ $\endgroup$ – Tacet Nov 6 '14 at 15:43
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Your last expression simplifies to

$${x+1\over\sqrt{x^2+x+1}}$$

There are now various ways to proceed. One is to divide numerator and denominator by $x$, bringing the $x$ in the denominator inside as an $x^2$:

$${x+1\over\sqrt{x^2+x+1}}={1+{1\over x}\over\sqrt{1+{1\over x}+{1\over x^2}}}$$

Do you see where this is going?

Added later: I answered too quickly, without checking how you got to the last expression. It turns out it's missing a "$+x$" in the denominator: it should have been

$${x+1\over\sqrt{x^2+x+1}+x}$$

so when you do the trick I advised you get

$${x+1\over\sqrt{x^2+x+1}}={1+{1\over x}\over\sqrt{1+{1\over x}+{1\over x^2}}+1}$$

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  • $\begingroup$ Yes I believe; $$\frac{1+0}{\sqrt{1+0+0}}=1$$ right?. But then is my asymptote: $ax+b=x+1?$ Because that's what i keep coming up with and it's not correct somehow... $\endgroup$ – Strange Brew Nov 6 '14 at 15:42
  • $\begingroup$ Yes! Thank you! That was my error all along! $\endgroup$ – Strange Brew Nov 6 '14 at 16:04
  • $\begingroup$ @StrangeBrew, it's often easier to spot the error someone else made than it is to spot your own. $\endgroup$ – Barry Cipra Nov 6 '14 at 16:06
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Don't do the difference, divide!

$\lim_{x->\infty}\dfrac{\sqrt{x^2+x+1}}{x}=1$ very easily

For $\dfrac{\sqrt{x^2+x+1}}{x}=\dfrac{\sqrt{x^2+x+1}}{\sqrt{x^2}}=\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x²}}$

EDIT: It is not over, of course, since the asymptote does not necessarily goes through the origin... You have to make the actual difference between the function and $x+a$. But by using the same strategy, you will find easily the result.

EDIT2: in order to be more complete.

$\sqrt{x^2+x+1}-x=x\times(\sqrt{1+\frac{1}{x}+\frac{1}{x²}}-1)=\dfrac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x²}}+1}$

Whose limit is clearly $\dfrac{1}{2}$.

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  • $\begingroup$ I'm not sure I follow your logic. By dividing by x don't we lose the b, since it will go to 0 when we take the limit? Isn't this just the 1 i have as my coefficient? I'm trying to follow my textbook which explains finding the oblique asymptote by $$ lim_{x->\infty} f(x)-(ax+b)$$ Where f(x) goes to 0 when we take the limit. Thus i get the coefficient a by: $$ lim_{x->\infty} \frac{f(x)}{x}-a $$ But I can't divide by x when I'm trying to figure out b. $\endgroup$ – Strange Brew Nov 6 '14 at 15:54
  • $\begingroup$ That is the second step that I explained afterwards. $\endgroup$ – Martigan Nov 6 '14 at 15:55
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$$\lim_{x->\infty}\sqrt{x^2+x+1}-x$$ $$=\lim_{x->\infty}\left(\sqrt{x^2+x+1}-x\right)\cdot\frac{\sqrt{x^2+x+1}+x}{\sqrt{x^2+x+1}+x}$$ $$=\lim_{x->\infty}\frac{1+x}{\sqrt{x^2+x+1}+x}$$ The limit of a sum is the sum of the limits $$=\lim_{x->\infty}\frac{1}{\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{x}{\sqrt{x^2+x+1}+x}$$ The limit of a quotient is the quotient of the limits
The limit of a constant is the constant $$=\frac{1}{\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{x}{\sqrt{x^2+x+1}+x}$$ $$=\frac{1}{\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$ The limit of a sum is the sum of the limits $$=\frac{1}{\infty+\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$ Use the power law $$=\frac{1}{\infty+\sqrt{\displaystyle\lim_{x->\infty}x^2+x+1}} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$ $$=\frac{1}{\infty+\infty} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$ $$=\frac{1}{\infty} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$ $$=0 + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$ $$=\lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$ The limit of a quotient is the quotient of the limits
The limit of a constant is the constant
The limit of a sum is the sum of the limits $$=\frac{1}{1+\displaystyle\lim_{x->\infty}\frac{\sqrt{x^2+x+1}}x}$$ $$=\frac{1}{1+\displaystyle\lim_{x->\infty}\sqrt{\frac{x^2+x+1}{x^2}}}$$ Use the power law $$=\frac{1}{1+\sqrt{\displaystyle\lim_{x->\infty}\frac{x^2+x+1}{x^2}}}$$ Use L'Hopital $$=\frac{1}{1+\sqrt{\displaystyle\lim_{x->\infty}1+\frac{1}{2x}}}$$ $$=\frac{1}{1+\sqrt{\displaystyle1+\frac12\cdot\lim_{x->\infty}\frac{1}{x}}}$$ $$=\frac{1}{1+\sqrt{\displaystyle1+\frac12\cdot0}}$$ $$=\color{lightgray}{\boxed{\color{black}{\dfrac{1}{2}}}}$$

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  • $\begingroup$ You copied this or what? $\endgroup$ – user190024 Nov 6 '14 at 15:53
  • $\begingroup$ @Infinity It's simple application of simple rules. $\endgroup$ – Alice Ryhl Nov 6 '14 at 15:56
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Easiest way is to complete to square $$\lim_{x\to\infty}\sqrt{x^2+x+1}-x=\sqrt{(x+1/2)^2+3/4}-x=\sqrt{(x+1/2)^2}-x=\sqrt{(x+1/2)^2(1+\frac{3}{4(x+1/2)^2})}-x=(x+1/2)-x=1/2$$

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  • $\begingroup$ That does seem very neat. I'm wondering how you got from step 2 to step 3 and also from step 3 to step 4. Maybe you could direct me to some more general explanation? $\endgroup$ – Strange Brew Nov 6 '14 at 17:11
  • $\begingroup$ @StrangeBrew I made i mistake It should be Step 2 to Step 4,and Step 4 to Step 3 to Step 5.Anyway I've rewriten $\sqrt{(x+1/2)^2+3/4}$ from the brackets so it became $\sqrt{(x+1/2)^2(1+\frac{3}{4(x+1/2)^2}}$ now I've wrote it as $\sqrt{(x+1/2)^2}\sqrt{1+\frac{3}{4(x+1/2)^2}}$.And the second root $\sqrt{1+\frac{3}{4(x+1/2)^2}}$ equal to $1$ since $x$ tends to infinity so $\frac{3}{4(x+1/2)^2}=0$. $\endgroup$ – kingW3 Nov 6 '14 at 17:38
  • $\begingroup$ Okay that explains a bit more. However I still don't understand quite what you've done. The way i calculate it; $$\sqrt{(x+1/2)^2+3/4}=\sqrt{(x+1/2)^2+\frac{3(x+1/2)^2}{4(x+1/2)^2}}=\sqrt{(x+1/2)^2\frac{3x^2+3x+1/4}{4(x+1/2)^2}}=?$$ Here I don't see the continuation to get to your expression. $\endgroup$ – Strange Brew Nov 6 '14 at 18:02
  • $\begingroup$ Well $(x+1/2)^2(1+\frac{3}{4(x+1/2)^2})=(x+1/2)^2+\frac{3(x+1/2)^2}{4(x+1/2)^2}$ $\endgroup$ – kingW3 Nov 6 '14 at 21:18
  • $\begingroup$ Okay, how do you get there? $\endgroup$ – Strange Brew Nov 9 '14 at 3:33
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We have $$\sqrt{x^2 + x + 1} - x = \frac{x^2 + x + 1 - x^2}{\sqrt{x^2 + x + 1} + x} = \frac{x + 1}{\sqrt{x^2 + x + 1} - x}.$$ L'Hopital's rule then gives $$\lim_{x\to\infty} \sqrt{x^2 + x + 1} - x = \lim_{x\to\infty}\left(1 + \frac{2x+1}{2\sqrt{x^2 + x + 1}}\right)^{-1} = \frac{1}{2}.$$

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I think there's a mistake in your last expression. We have $$ \sqrt{x^2+x+1}-x = \frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x} = \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} $$

Therefore we have $$\lim(\sqrt{x^2+x+1}-x)= \lim\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} = \frac{1}{2} $$

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  • $\begingroup$ I honestly don't know why. $\endgroup$ – primitiveroot Nov 6 '14 at 15:54
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    $\begingroup$ You forgot a bracket or two. :) $\endgroup$ – Richard D. James Nov 6 '14 at 15:55
  • $\begingroup$ @SpamIAm, actually there was one too many. I removed it. $\endgroup$ – Barry Cipra Nov 6 '14 at 16:11

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