# Limit problem (what's going wrong?)

• Jun 16th 2013, 12:23 AM
HeilKing
Limit problem (what's going wrong?)
$\displaystyle {\lim _{x \to - \infty }}{\frac{4x}{\sqrt{x^{2}+4x}+x}}}$

This gives $\displaystyle - \infty$, which is correct.

Dividing both the numerator and the denominator by $\displaystyle x$, we have

$\displaystyle {\lim _{x \to - \infty }}{\frac{4}{\sqrt{1+\frac{4}{x}}+1}}}$

now this gives me 2.

What's wrong?

P.S. I tested the values on a calculator.
• Jun 16th 2013, 02:00 AM
ibdutt
Re: Limit problem (what's going wrong?)
Why do you feel it is wrong and what do you think is wrong? have confidence in yourself.
• Jun 16th 2013, 02:28 AM
HeilKing
Re: Limit problem (what's going wrong?)
I mean they even become two different functions after the division. Their behaviors differ when x<-4.
• Jun 16th 2013, 03:22 AM
Plato
Re: Limit problem (what's going wrong?)
Quote:

Originally Posted by HeilKing
$\displaystyle {\lim _{x \to - \infty }}{\frac{4x}{\sqrt{x^{2}+4x}+x}}}$
This gives $\displaystyle - \infty$, which is correct.
Dividing both the numerator and the denominator by $\displaystyle x$, we have
$\displaystyle {\lim _{x \to - \infty }}{\frac{4}{\sqrt{1+\frac{4}{x}}+1}}}$
now this gives me 2.
What's wrong?.

If $\displaystyle x<0$ then $\displaystyle {\frac{4x}{\sqrt{x^{2}+4x}+x}}<0$. So what is the limit?
• Jun 16th 2013, 02:16 PM
BobP
Re: Limit problem (what's going wrong?)
I think that the first answer is incorrect. I think that the value of the limit is $\displaystyle +\infty$, not $\displaystyle -\infty$.

If $\displaystyle x(<-4)$ is negative then $\displaystyle \sqrt{x^{2}+4x}$ is less than $\displaystyle x$ in magnitude which implies that the denominator is negative and the fraction positive.

There is a presumption in the notation that $\displaystyle \sqrt{x^{2}+4x}$ is a positive quantity. What is happening when the $\displaystyle x$ is cancelled is as if a $\displaystyle -\infty$ is coming out from the square root in effect making it negative. (That is, $\displaystyle \sqrt{x^2+4x}$ tends to $\displaystyle -\infty$ rather than $\displaystyle +\infty$ as it should).

A way round the problem is to replace $\displaystyle x$ by $\displaystyle -x$ and take the limit as $\displaystyle x \rightarrow +\infty.$
• Jun 16th 2013, 04:30 PM
HallsofIvy
Re: Limit problem (what's going wrong?)
In the first form, if you are "taking x going to $\displaystyle -\infty$", the numerator goes to $\displaystyle -\infty$ goes to $\displaystyle -\infty$ while the denominator goes to $\displaystyle \infty- \infty$, an indeterminant form. Notice that if x= -10000, then $\displaystyle \frac{4x}{\sqrt{x^2+ 4x}+ x}= \frac{-40000}{\sqrt{9999}- 10000}= \frac{-40000}{9999.5}= -4.0002$ which is NOT going to "$\displaystyle -\infty$".
• Jun 16th 2013, 05:57 PM
johng
Re: Limit problem (what's going wrong?)
Hi,
BobP is correct. The limit is $\displaystyle +\infty$.
The problem is some subtle elementary algebra. For x < 0,$\displaystyle \sqrt{x^2}=-x$. So for x < 0,
$\displaystyle {\sqrt{x^2+4x}\over x}={\sqrt{x^2+4x}\over-\sqrt{x^2}}=-{\sqrt{{x^2+4x\over x^2}}=-\sqrt{1+{4\over x}}$
Thus for x < 0, $\displaystyle {4x\over\sqrt{x^2+4x}+x}={4\over-\sqrt{1+{4\over x}}+1$
Since x < 0, the denominator is positive and approaches 0 as x goes to infinity. So the original limit is plus infinity.
• Jun 16th 2013, 07:59 PM
topsquark
Re: Limit problem (what's going wrong?)
Quote:

Originally Posted by HallsofIvy
Notice that if x= -10000, then $\displaystyle \frac{4x}{\sqrt{x^2+ 4x}+ x}= \frac{-40000}{\sqrt{9999}- 10000}= \frac{-40000}{9999.5}= -4.0002$ which is NOT going to "$\displaystyle -\infty$".

A correction here, maybe a typo, but it changes a lot.
$\displaystyle \frac{4x}{\sqrt{x^2+ 4x}+ x}= \frac{-40000}{9999 - 10000}= \frac{-40000}{-1}= 40000$

Let's try this for a larger x, say x = -10^9. Then
$\displaystyle \frac{4x}{\sqrt{x^2+ 4x}+ x} = \frac{-4 \times 10^9}{ 10^9 -10^9 }$
The denominator here isn't really 0, just that the number from the square root is very close to 10^9, and smaller than the 10^9 from the x term in the denominator...it also makes the denominator negative. So the expression is very much larger than +10^9. Sounds like it's headed for infinity to me.

-Dan
• Jun 17th 2013, 06:18 AM
mnov
Re: Limit problem (what's going wrong?)
Since $\displaystyle x$ is going to $\displaystyle -\infty$, write $\displaystyle x$ as $\displaystyle -|x|$. The given limit is
$\displaystyle \lim_{|x|\to\infty} \frac{-4|x|}{\sqrt{|x|^2 -4|x|} -|x|}$
$\displaystyle = \lim_{|x|\to\infty} \frac{-4}{\sqrt{1 -\frac{4}{|x|}} -1}$
$\displaystyle = \lim_{\epsilon\downarrow 0} \frac{4}{1 - \sqrt{1 -\epsilon} } = +\infty$
• Jun 18th 2013, 09:14 PM
mpx86
Re: Limit problem (what's going wrong?)
first it can be written as

$\displaystyle $\mathop {\lim }\limits_{x \to - \infty } \frac{{4x}}{{\sqrt {{x^2} + 4x} + x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{4x}}{{\sqrt {{x^2} + 4x} + x}}*(\sqrt {{x^2} + 4x} - x)/(\sqrt {{x^2} + 4x} - x) = \mathop {\lim }\limits_{x \to - \infty } \frac{{4x}}{{4x}}*(\sqrt {{x^2} + 4x} - x) = \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2} + 4x} - x$$

now you cannot take x out since it is a negative number

this question can be done in the following way

since x is negative , convert it to a positive number by substituting y= -x or x= -y

therefore functon becomes $\displaystyle $\mathop {\lim }\limits_{x \to - \infty } (\sqrt {{x^2} + 4x} - x) = \mathop {\lim }\limits_{y \to \infty } \sqrt {{y^2} - 4y} + y = \mathop {\lim }\limits_{y \to \infty } y(\sqrt {1 - 4/y} + 1) = \mathop {\lim }\limits_{y \to \infty } y(1 + 1) = \mathop {\lim }\limits_{y \to \infty } 2y = \infty$$

thus answer is infinity not minus infinity......