Can anyone please help me with this task?

[Nadia_91]

Hello I have a task which I never have done before, can anybody please help me?

The task:

A) Find f'(x) by calculation using the rules for differention.

\(\displaystyle f(x)=2x^{2}-3x\)

B) Find the limit.

\(\displaystyle \lim_{x \to -2}\frac{x^{2}-4}{3x+6}\)

I need full calculation in order so I can understand this task please?

 

[Aladdin]

-------------------Hey Nadia ,
f(x) = 2x^2 -3x

f'(x) = 2(2)x^(2-1) -3(1)x^(1-1) = 4x -3.

Lim ( x^2 - 4)/( 3x +6) as x tends to -2 equals ; 0/0 indeterminate , use Hopital's 'Rule

lim (2x / 3) as x tends to -2 ; => ; -4/3.

 

[Nadia_91]

Thank you for your help

But is this task entirely completed with full calculation?

 

[lookagain]

B) Find the limit.

\(\displaystyle \lim_{x \to -2}\frac{x^{2}-4}{3x+6}\)

I need full calculation in order so I can understand this task please?

Nadia_91,

I'm not going to give you a full calculation, because you are to show some work.


Try an alternate method for this one:

\(\displaystyle \ \ \lim_{x \to \ -2} \ \frac{x^2 - 4}{3x + 6} \ \)

Factor the numerator and the denominator:

\(\displaystyle = \ \lim_{x \to \ -2} \ \frac{(x - 2)(x + 2)}{3(x + 2)}\)

Cancel the factors, and you can do that because x never actually takes on a value of -2:

\(\displaystyle = \ \lim_{x \to \ -2} \ \frac{x - 2}{3}\)


Now you can finish this. . .