# Limit at infinity equaling infinity?

• October 1st 2010, 07:11 PM
DannyMath
Limit at infinity equaling infinity?
Hi guys,

I'm doing limit exercises and I got this question:

Evaluate the following limit:

lim
x → −∞ √(x2 + 5 x + 1) - x

After working it out by multiplying by conjugate I get 5/0 which is undefined, but I know the answer is infinity. So where did I go wrong? This is the last step which I deduced 5/0 from:

(5 + (0))/-sqrt(1 + (0) + (0)) + 1

Any clarification is appreciated :)
• October 1st 2010, 09:10 PM
Soroban
Hello, DannyMath!

Quote:

$\displaystyle \text{Evaluate: }\;\lim_{x\to-\infty}\sqrt{x^2+5x+1} - x$

$\text{. . . but I know the answer is infinity.}$ .?

Multiply numerator and denominator by the conjugate:

. . $\displaystyle \frac{\sqrt{x^2+5x+1} - x}{1}\cdot\frac{\sqrt{x^2+5x+1} + x}{\sqrt{x^2+5x+1} + x}$

. . $\displaystyle =\;\frac{(x^2+5x +1) - x^2}{\sqrt{x^2+5x+1} + x} \;=\; \frac{5x+1}{\sqrt{x^2+5x+1} + x}$

$\displaystyle \text{Divide numerator and denominator by }x\!:\;\; \displaystyle \frac{\dfrac{5x+1}{x}}{\dfrac{\sqrt{x^2+5x+1} + x}{x}}\;\;\bf[1]$

$\displaystyle \text{The numerator is: }\;\frac{5x}{x} + \frac{1}{x} \;=\;5 + \frac{1}{x}$

$\displaystyle \text{The denominator is: }\;\frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}$

. . $\displaystyle \;=\;\frac{\sqrt{x^2+5x+1}}{\sqrt{x^2}} + 1\;=\;\sqrt{\frac{x^2+5x+1}{x^2}} + 1$

. . $\displaystyle =\;\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1$

$\displaystyle \text{Then }{\bf[1]}\text{ becomes: }\; \frac{5 + \frac{1}{x}}{\sqrt{1+\frac{5}{x} + \frac{1}{x^2}} + 1}$

$\displaystyle \text{Therefore: }\;\lim_{x\to-\infty}\frac{5+\frac{1}{x}}{\sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1} \;=\;\frac{5 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{5}{2}$

• October 1st 2010, 09:15 PM
DannyMath
Hey, thanks for the response! This was actually a 2 part exercise I was given for homework and I only posted the part (b). Part (a) was the exact same question but x goes to infinity rather than negative infinity. For that one I got 5/2 and the system told me I was right. But for the one I posted I got 5/0 but my graphing program showed it going to no particular horizontal asymptote so I had a sneaking suspicion that it was infinity. So that's what I submitted and it said I was correct :S ??

Also, in the denominator, since you're turning x into sqrt(x^2), aren't you losing a negative sign if you don't write it -sqrt(x^2)? Since x would be negative as x >> infinity?
• October 1st 2010, 10:25 PM
tonio
Quote:

Originally Posted by Soroban
Hello, DannyMath!

Multiply numerator and denominator by the conjugate:

. . $\displaystyle \frac{\sqrt{x^2+5x+1} - x}{1}\cdot\frac{\sqrt{x^2+5x+1} + x}{\sqrt{x^2+5x+1} + x}$

. . $\displaystyle =\;\frac{(x^2+5x +1) - x^2}{\sqrt{x^2+5x+1} + x} \;=\; \frac{5x+1}{\sqrt{x^2+5x+1} + x}$

$\displaystyle \text{Divide numerator and denominator by }x\!:\;\; \displaystyle \frac{\dfrac{5x+1}{x}}{\dfrac{\sqrt{x^2+5x+1} + x}{x}}\;\;\bf[1]$

$\displaystyle \text{The numerator is: }\;\frac{5x}{x} + \frac{1}{x} \;=\;5 + \frac{1}{x}$

$\displaystyle \text{The denominator is: }\;\frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}$

. . $\displaystyle \;=\;\frac{\sqrt{x^2+5x+1}}{\sqrt{x^2}} + 1\;=\;\sqrt{\frac{x^2+5x+1}{x^2}} + 1$

You can't do the above since $x<0$...it can't go into the square root. What has to be done is:

$\frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}=-\frac{\sqrt{x^2+5x+1}}{-x} + \frac{x}{x}=-\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=-\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1$...etc.

At the end the limit indeed is $\infty$

Tonio

. . $\displaystyle =\;\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1$

$\displaystyle \text{Then }{\bf[1]}\text{ becomes: }\; \frac{5 + \frac{1}{x}}{\sqrt{1+\frac{5}{x} + \frac{1}{x^2}} + 1}$

$\displaystyle \text{Therefore: }\;\lim_{x\to-\infty}\frac{5+\frac{1}{x}}{\sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1} \;=\;\frac{5 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{5}{2}$

.
• October 1st 2010, 10:53 PM
DannyMath
Thank you Tonio, I did indeed get that but doesn't that end up being (5 + (0))/-sqrt(1 + (0) + (0)) + 1 which turns into 5/0? I've just started limits so I don't know if that is the correct way to finish the problem (obviously it's not since it gives wrong answer :P)
• October 1st 2010, 11:00 PM
tonio
Quote:

Originally Posted by DannyMath
Thank you Tonio, I did indeed get that but doesn't that end up being (5 + (0))/-sqrt(1 + (0) + (0)) + 1 which turns into 5/0? I've just started limits so I don't know if that is the correct way to finish the problem (obviously it's not since it gives wrong answer :P)

Well, as you know you can't divide by zero...(Cool)

What you actually get is an expression of the form $\frac{c}{g(x)}$ , with $c$ a constant and $g(x)$ a

function which converges to zero. It's not hard to see, even by means of the $\epsilon,\,\delta$ definition, that the limit of

this expression has as limit $\pm\infty$ , depending on the constant's and the function's signs.

Tonio
• October 1st 2010, 11:07 PM
DannyMath
Oh ok I will work on understanding that. We skipped the precise delta/epsilon description in our books and were just told to divide by greatest x degree in denominator. I just tried to go as far as I could with the definition. Maybe one day it will be as easy to see for me as it is for you guys! HAHA! Thanks!