Derivatives of functions and limits, please help!

If f is a differentiable function and f' is continuous with f(2)=0 and f'(2)=7, find:

lim x->0 (f(2+3x)+f(2+5x))/x

http://imageshack.us/photo/my-images/808/picture4whx.png/

What does it mean for f' to be continuous? And do you apply the quotient rule here? Thanks!

This is a L'Hopital's rule question. Note that when x=0, the numerator is equal to
f(2+3*0)+f(2+5*0)=
=f(2+0)+f(2+0)=
=f(2)+f(2)=
=0+0=
=0,
and the denominator is 0 also. So we have the indeterminant form 0/0. L'Hopital says to consider the ratio of the derivatives (which is *not* the same as using the quotient rule.)

The derivative of the top is f '(2+3x)*3+f '(2+5x)*5 [use the chain rule for each term.]

The derivative of the bottom is just 1.

L'Hopital says that *if*
lim x-->0 [(f '(2+3x)*3+f '(2+5x)*5)/1] exists, then the original limit exists too, and has the same value.

To say that f ' is continuous means (for all intent and purposes) that we can evaluate limits involving f ' by plugging in for x.

So lim x-->0 [(f '(2+3x)*3+f '(2+5x)*5)/1]=
=f '(2+3*0)*3+f '(2+5*0)*5=
=f '(2)*3+f '(2)*5=
=7*3+7*5=
=7*8=
=56.

By L'Hopital's rule, lim x->0 (f(2+3x)+f(2+5x))/x exists and is equal to 56.

The definition of continuity is a little complicated, but here it basically just means f' behaves nicely so you can do all your usual stuff with it. Now, as x goes to 0, f(2 + 3x) must also go to 0, because 2 + 3x is going to 2 and f(2) = 0. Similarly, f(2 + 5x) is going to 0. Obviously, x (in the denominator) is also going to 0. Since everything is going to 0, that means this is a good time to break out l'hopital's rule! Taking the derivative of the top and the bottom, we now want to find lim x->0 of f'(2x + 3) + f'(2 + 5x), and the denominator goes away because the derivative of x is 1. Now, the limit of each piece is 7, since 2 + 3x and 2 + 5x each go to 2 as x goes to 0, and f'(2) = 7. Since the lim x->0 of 14 is just 14, that is your answer. No quotient rule needed.

Hope this helps!

Since f(2) = 0, let's write:

lim x-> 0 [ ( f(2+3*x) - f(2) + f(2+5*x) - f(2) ) /x ]

The first two terms give:

lim x-> 0 [ ( f(2+3*x) - f(2) )/ x ]
=lim x-> 0 [ 3*( f(2+3*x) - f(2) )/(3*x) ]
=lim u-> 0 [ 3*( f(2+u) - f(2) )/u ]
= 3*f '(2)

In the third line, I used the substitution u =3*x. Similarly, let u = 5*x to evaluate the second pair of terms, to get 5*f '(2). Thus,

lim x-> 0 [ ( f(2+3*x) + f(2+5*x) ) /x ]
=3*f '(2) + 5*f '(2)
=8*7 = 56

I believe continuity of the derivative is required to evaluate the limit this way.