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.

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.

Re: Limit problem (what's going wrong?)

I mean they even become two different functions after the division. Their behaviors differ when x<-4.

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?.

Put your calculator away. Use your brain.

If $\displaystyle x<0$ then $\displaystyle {\frac{4x}{\sqrt{x^{2}+4x}+x}}<0$. So what is the limit?

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 $\displaystyle x \rightarrow +\infty.$

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 "$\displaystyle -\infty$".

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.

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

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

$\displaystyle \lim_{|x|\to\infty} \frac{-4|x|}{\sqrt{|x|^2 -4|x|} -|x|}$

$\displaystyle = \lim_{|x|\to\infty} \frac{-4}{\sqrt{1 -\frac{4}{|x|}} -1}$

$\displaystyle = \lim_{\epsilon\downarrow 0} \frac{4}{1 - \sqrt{1 -\epsilon} } = +\infty$

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......