Question

Find the limit if f(x) is differentiable at x=a

Answer 1

\[\large \lim_{x \rightarrow a} \dfrac{x^2f(a)-a^2f(x)}{x-a}\]

Answer 2

Does l'hopital work? I haven't tried

Answer 3

But we do not know if this is 0/0 or infinity/infinity form+ we do not values of f(a) and f'(a).

Answer 4

Oh yeah. What about values of f(a) and f'(a)

Answer 5

But f(x) is differentiatable at x=a

Answer 6

L hospital worked :)

I got 2af(a)-a^2f'(a)

Answer 7

You mean f'(x) instead of f'(a)

Never mind you substituted it

Answer 8

Yeah..It turns out I didnt need the values of f(a) and f'(a) I checked it with the solution. Thanks!

Answer 9

no problem; I learnt something too :)

Answer 10

I believe if you let h = x - a would do the trick

Answer 11

I'm intrigued.

Answer 12

substitute every we have:
(a + h)^2 f(a) - a^2 f(a + h)
---------------------------
h

Answer 13

What about a^2f(a+h)/h?

Answer 14

now expand everything:
(a^2 + 2ah + h^2) f(a) - a^2 f(a + h)
--------------------------------
h
(a^2/h + 2a + h) f(a) - a^2 * f(a + h)/h

distribute:
f(a) (2a + h) + f(a) (a^2/h) - a^2 * f(a + h)/h

f(a) (2a + h) + a^2 [ f(a)/h - f(a+h)/h ]

f(a) (2a + h) - a^2 [f(a+h) - f(a)] /h

as h approaches 0, the last term is just f'(a), so we have
f(a) (2a + 0) - a^2 f'(a)

f(a) (2a) - a^2 f'(a) <---

Answer 15

A clever method; but much more complicated? I don't know o.o

Answer 16

well I wouldn't say complicated since it's was just algebra. But clever, yes :DD

Answer 17

I have seen this manipulation before. Thanks for explaining it here @sourwing.

Answer 18

yw ;)