This gives , which is correct.

Dividing both the numerator and the denominator by , we have

now this gives me 2.

What's wrong?

P.S. I tested the values on a calculator.

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- Jun 16th 2013, 12:23 AMHeilKingLimit problem (what's going wrong?)

This gives , which is correct.

Dividing both the numerator and the denominator by , we have

now this gives me 2.

What's wrong?

P.S. I tested the values on a calculator. - Jun 16th 2013, 02:00 AMibduttRe: 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 AMHeilKingRe: 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 AMPlatoRe: Limit problem (what's going wrong?)
- Jun 16th 2013, 02:16 PMBobPRe: Limit problem (what's going wrong?)
I think that the first answer is incorrect. I think that the value of the limit is , not .

If is negative then is less than in magnitude which implies that the denominator is negative and the fraction positive.

There is a presumption in the notation that is a positive quantity. What is happening when the is cancelled is as if a is coming out from the square root in effect making it negative. (That is, tends to rather than as it should).

A way round the problem is to replace by and take the limit as - Jun 16th 2013, 04:30 PMHallsofIvyRe: Limit problem (what's going wrong?)
In the first form, if you are "taking x going to ", the numerator goes to goes to while the denominator goes to , an

**indeterminant**form. Notice that if x= -10000, then which is NOT going to " ". - Jun 16th 2013, 05:57 PMjohngRe: Limit problem (what's going wrong?)
Hi,

BobP is correct. The limit is .

The problem is some subtle elementary algebra. For x < 0, . So for x < 0,

Thus for x < 0,

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 PMtopsquarkRe: Limit problem (what's going wrong?)
A correction here, maybe a typo, but it changes a lot.

Let's try this for a larger x, say x = -10^9. Then

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 AMmnovRe: Limit problem (what's going wrong?)
Since is going to , write as . The given limit is

- Jun 18th 2013, 09:14 PMmpx86Re: Limit problem (what's going wrong?)
first it can be written as

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

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