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I have the equation below:

$$ lim_{x\to \infty} \frac{\sqrt {4x^2+4x}}{4x+1} $$

The hint provided to solve this problem is as below:

To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit.

Since the leading term of the denominator is $x$, let's divide by $x$.

In the numerator, let's divide by $\sqrt {x^2}$, since for positive values, $x = \sqrt {x^2}$.

$$ lim_{x\to \infty} \frac{\sqrt {4x^2+4x}}{4x+1} $$ $$ lim_{x\to \infty} \frac {\frac{\sqrt {4x^2+4x}}{\sqrt {x^2}}} {\frac{4x+1}{x} } $$

Can you show me how the values $x = \sqrt {x^2}$? I'd like to see the proof. Thank you.

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    $\begingroup$ It’s a typo: it’s supposed to read $x=\sqrt{x^2}$. $\endgroup$
    – Brian M. Scott
    Nov 21, 2016 at 0:15
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    $\begingroup$ That looks like a typo. I think they mean to say "In the numerator, let's divide by $\sqrt{x^2}$, since for positive $x$, we have $x=\sqrt{x^2}$" $\endgroup$
    – Arthur
    Nov 21, 2016 at 0:15
  • $\begingroup$ The typo was on my end, I've fixed it now. I still have the same question though, why does $x = \sqrt {x^2}$? $\endgroup$
    – supmethods
    Nov 21, 2016 at 1:21

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In general for $x\in\mathbb R$, you have that $$\sqrt{x^2} = |x|,$$ but $|x|=x$ whenever $x$ is positive. Since the limit is for $x\to \infty$ we can assume $x>0$ and hence it follows $$\sqrt{x^2} = x.$$

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  • $\begingroup$ Thanks. I have one other questions about conventions for $\infty$. In general if there is no sign for $\infty$, will this mean both negative and positive $\infty$? That is: ${x\to \infty}$ means infinity in both directions, positive infinity ${x\to \infty^+}$ is infinity in the positive direction, negative infinity ${x\to \infty^-}$ is infinity in the negative direction. $\endgroup$
    – supmethods
    Nov 21, 2016 at 1:58
  • $\begingroup$ Generally, the expression $\lim\limits_{x\to \infty} f(x)$ is the limit of $x$ going to positive infinity. However, rarely I have seen cases where it simply means that $x$ get unboundedly large in either direction. It should usually be understandable from the context. $\endgroup$
    – Eff
    Nov 21, 2016 at 2:03
  • $\begingroup$ In real analysis (as opposed to complex), $\infty$ is the same as $+\infty$, which is of course different from $-\infty$. Not that the signs should go in front of the infinity symbol; writing $\infty^{\pm}$ doesn't make sense. $\endgroup$
    – Hans Lundmark
    Nov 21, 2016 at 6:21

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