How do you solve this tough Pre-Calculus Problem?

Question

Let f be defined in the real numbers as f(x) = 3x^2-2x+5. Find the lim f(x)-f(a)/(x-a) when x --> a.

Answer 1

lim (3x^2-2x+5-(3a^2-2a+5))/(x-a) when x --> a
lim (3(x^2-a^2)-2(x-a))/(x-a) when x --> a
lim (3(x-a)(x+a)-2(x-a))/(x-a) when x --> a
lim (x-a)(3(x+a)-2))/(x-a) when x --> a
lim 3(x+a)-2 when x --> a
3(a+a)-2
3(2a)-2
6a-2

Answer 2

Method 1: Note that this is simply another way of writing the derivative. Conclude that this is f'(x) evaluated at a, which is 6a-2

Method 2: since direct substitution gives you 0/0, use L'hopitals rule and find that [x→a]lim (f(x)-f(a))/(x-a) = [x→a]lim f'(x)/1 = f'(a) = 6a-2 (essentially the same as above, but using a rule rather than the similarity to the difference quotient)

Method 3: Direct approach:

[x→a]lim (3x² - 2x + 5 - (3a² - 2a + 5))/(x-a)
Simplify:
[x→a]lim (3x² - 3a² - 2x + 2a)/(x-a)
Split the numerator across two fractions:
[x→a]lim (3x² - 3a²)/(x-a) - (2x - 2a)/(x-a)
Factor the numerators:
[x→a]lim 3(x+a)(x - a)/(x-a) - 2(x-a)/(x-a)
Cancel:
[x→a]lim 3(x+a) - 2
Evaluate:
3(a+a)-2
6a-2

Answer 3

Just let x=a+h

plug into the expression, expand, and let h-> 0 to get the

f ' (a) = 6a-2, the derivative of f(x) at x=a

Answer 4

Let's replace x with (a + da). This then becomes the limit of

f(a+da)-f(a) / da

Plugging into the equation...

3(a+da)^2 - 2(a+da) + 5 - 3a^2 + 2a -5
-------------------------------------------------
da

The numerator simplifies to 6a da + 3 da^2 -2 da and when you divide through by the denominator you get

6a + 3 da -2

The limit of this as da->0 is just 6a-2.