This gives , which is correct.
Dividing both the numerator and the denominator by , we have
now this gives me 2.
P.S. I tested the values on a calculator.
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
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.
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.
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......