Understanding Infinity: Positive vs. Negative Infinity Explained

[lastochka]

View attachment 3871
Hello,
sorry I tried to use Latex, but it didn't work...I uploaded picture of what I did instead.
I have a small question about the answer which is infinity why this is positive infinity? I know that correct answer is positive infinity, but I am trying to find explanation why...How do we know it is positive and not negative? Is it only because we are coming from the right side? I am very confused with this question. Also in which case there is negative infinity?
Thank you so much

 

[ineedhelpnow]

Hello. Remember to enclose you latex with dollar signs or with the math tag.
So with regards to your question, take the limit of $\sqrt{x^2+7x+1}$ as x goes to $ -\infty$ and the same for x. You will find that the first is equal to $ \infty$ and the second to $- \infty$. Now go back to the original question. You have $\sqrt{x^2+7x+1}-x$. Plug in the solutions. $\infty - (- \infty)$ is equal to $\infty$. I hope that helps. That's the mathematical part. As for the conceptual part, I'll leave that to someone else who can explain better than me. :)

 

[MarkFL]

Another method to consider is to make the substitution:

\(\displaystyle u=-\frac{1}{x}\)

and our limit becomes:

\(\displaystyle \lim_{u\to0}\frac{\sqrt{u^2-7u+1}+1}{u}=\infty\)

 

[lastochka]

Hello. Remember to enclose you latex with dollar signs or with the math tag.
So with regards to your question, take the limit of $\sqrt{x^2+7x+1}$ as x goes to $ -\infty$ and the same for x. You will find that the first is equal to $ \infty$ and the second to $- \infty$. Now go back to the original question. You have $\sqrt{x^2+7x+1}-x$. Plug in the solutions. $\infty - (- \infty)$ is equal to $\infty$. I hope that helps. That's the mathematical part. As for the conceptual part, I'll leave that to someone else who can explain better than me. :)

Thank you so much! I was wandering what I am doing wrong with Latex lol

You said that x will go to negative infinity...is it because it has negative sign?

Thank you again for helping me. I know it is probably easy question, but I am very new to this...

 

[Dethrone]

What ineedhelpnow meant was that $\lim_{{x}\to{-\infty}}x=-\infty$. So (intuitively), $\lim_{{x}\to{-\infty}}\sqrt{x^2+7x+1}-x=\lim_{{x}\to{-\infty}}\sqrt{x^2+7x+1}-\lim_{{x}\to{-\infty}}x= \infty-(-\infty)=+\infty$.

Another method to consider is to make the substitution:

\(\displaystyle u=-\frac{1}{x}\)

Now, I just wanted to point something out here that was implied. Notice that for $u=-\frac{1}{x}$, $\lim_{{x}\to{-\infty}}-\frac{1}{x}=\lim_{{x}\to{\infty}}\frac{1}{x}$, which is tending to 0 from the positive side. Thus, in terms of $u$, we have $\lim_{{u}\to{0^+}}u$. So in fact, explicitely, we have $\lim_{u\to0^+}\frac{\sqrt{u^2-7u+1}+1}{u}$. Note that the top is always greater than 0, and $u>0$. A very large number over a very small positive number going to 0 is positive infinity.

 

[lastochka]

What ineedhelpnow meant was that $\lim_{{x}\to{-\infty}}x=-\infty$. So (intuitively), $\lim_{{x}\to{-\infty}}\sqrt{x^2+7x+1}-x=\lim_{{x}\to{-\infty}}\sqrt{x^2+7x+1}-\lim_{{x}\to{-\infty}}x= \infty-(-\infty)=+\infty$.Now, I just wanted to point something out here that was implied. Notice that for $u=-\frac{1}{x}$, $\lim_{{x}\to{-\infty}}-\frac{1}{x}=\lim_{{x}\to{\infty}}\frac{1}{x}$, which is tending to 0 from the positive side. Thus, in terms of $u$, we have $\lim_{{u}\to{0^+}}u$. So in fact, explicitely, we have $\lim_{u\to0^+}\frac{\sqrt{u^2-7u+1}+1}{u}$. Note that the top is always greater than 0, and $u>0$. A very large number over a very small positive number going to 0 is positive infinity.

Thank you so much, Rido12! It is all clear now!
Thank you, everyone for help!