Evaluate lim x/((x+2)^2 - 4) as x-> 0 or state that the limit does not exist.
My book said I can substitute 0 into the equation. I did that and came up with
____0_____
(0+2)^2 -4 which simplifies down to
___0___ which = _0_ which = _0_ = 0.
2^2 - 4 4-4 0
That's indeterminate, so I did it another way and factored out (x+2)^2 first. This is what I got:
_____x____ = ___x___
x^2+4x+4-4 x^2 + 4
And then I substituted 0 to get:
___0____ = _0_ = _0_ = 0.
0^2 +4*0 0+0 0
I'd like to know; is 0 my limit, or does a result of 0 determine that no limit exists? My book is not clear on this. Please let me know!
_x_________ =
x^2+4x+4-4
= ___x____
X^2 + 4 x
= ____x____
x(x+4)
= _____1____
(x+4)
which is 1/4 when x = 0
Oh....That would make more sense. Thanks for explaining that!
To evaluate the limit lim x/((x+2)^2 - 4) as x approaches 0, you correctly substituted 0 into the equation and simplified to get the indeterminate form of 0/0. To resolve this indeterminate form, you factored the denominator as (x+2)^2 - 4 = (x+2+2)(x+2-2) = (x+4)(x) and simplified the expression further.
Now, when you substitute 0 into the factored expression, you get 0/((0+4)(0)) = 0/0. However, this still remains an indeterminate form which means we cannot conclude the value of the limit from this approach alone.
In this case, we can try another method to evaluate the limit. We can use algebraic manipulation to simplify the expression further and then substitute the value of x=0. Let's proceed with this approach.
Starting with the original expression x/((x+2)^2 - 4), expand the denominator using the difference of squares formula:
(x+2)^2 - 4 = (x+2+2)(x+2-2) = (x+4)(x)
Rewriting the original expression:
x/((x+4)(x))
Now, we can remove the common factor of x from the numerator and denominator:
x/(x(x+4))
Simplifying, we find:
1/(x+4)
Now, taking the limit of this expression as x approaches 0, substitute x=0:
1/(0+4) = 1/4
Therefore, the limit of x/((x+2)^2 - 4) as x approaches 0 is 1/4.
To summarize, by simplifying the expression and then substituting x=0, we find that the limit exists and is equal to 1/4, not 0. The result of 0 you obtained previously did not determine that the limit does not exist, but rather was an intermediate step that led to the correct answer.