# Limit problem

terryds

## Homework Statement

If ##\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1##.
then f'(1) = ...

A. -1
B. -1/3
C. 1/3
D. 1
E. 2

## Homework Equations

[/B]
##f'(x) = \lim_{h->0}\frac{f(x+h)-f(x))}{h}##

[/B]

Homework Helper
Gold Member
2021 Award

## Homework Statement

If ##\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1##.
then f'(1) = ...

A. -1
B. -1/3
C. 1/3
D. 1
E. 2

## Homework Equations

[/B]
##f'(x) = \lim_{h->0}\frac{f(x+h)-f(x))}{h}##

## The Attempt at a Solution

[/B]

First, rewrite the definition of the derivative limit. There are always two alternative ways o wrtite a limit. And, I'm not talking about epsilon-delta.

Second, a hint: you need to learn to spot a composite function when you see one!

terryds
First, rewrite the definition of the derivative limit. There are always two alternative ways o wrtite a limit. And, I'm not talking about epsilon-delta.

Second, a hint: you need to learn to spot a composite function when you see one!

Do you mean the L' Hospital rule?
By using L'Hospital rule, I get
##\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1##
##\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1##
##\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1##
##3a^2 f'(a^3) = -a##
##f'(a^3) = \frac{-a}{3a^2}##
##f'(a^3) = -\frac{1}{3a}##
##f'(1^3) = -\frac{1}{3}##

Is it correct ?

Homework Helper
Gold Member
2021 Award
Do you mean the L' Hospital rule?
By using L'Hospital rule, I get
##\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1##
##\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1##
##\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1##
##3a^2 f'(a^3) = -a##
##f'(a^3) = \frac{-a}{3a^2}##
##f'(a^3) = -\frac{1}{3a}##
##f'(1^3) = -\frac{1}{3}##

Is it correct ?

An interesting solution! That's probably valid, but it's not what I meant. Note that:

##f'(a) \equiv \lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a} \equiv \lim_{h\rightarrow 0} \frac{f(a + h)-f(a)}{h}##

Those are the two equivalent ways of writing the definition of a derivative.

And, if you let ##g(x) = f(x^3)##, then what you were given was effectively ##\forall a, \ g'(a) = -1##. The rest follows from the chain rule.

terryds
An interesting solution! That's probably valid, but it's not what I meant. Note that:

##f'(a) \equiv \lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a} \equiv \lim_{h\rightarrow 0} \frac{f(a + h)-f(a)}{h}##

Those are the two equivalent ways of writing the definition of a derivative.

And, if you let ##g(x) = f(x^3)##, then what you were given was effectively ##\forall a, \ g'(a) = -1##. The rest follows from the chain rule.

g'(a) = -1
3a^2 f'(a^3) = -1
We let a = 1
f'(1)= -1/3
Thanks for your help

Mentor
As the problem is solved, here is a different approach: Define ##y=x^3##, ##b=a^3##.
Then
$$-1 = \lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} \\ = \lim_{y\rightarrow b} \frac{f(y)-f(b)}{\sqrt[3]{y}-\sqrt[3]{b}} \cdot \frac{\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2}{\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2} \\ = \lim_{y\rightarrow b} \frac{f(y)-f(b)}{y-b} \cdot \left(\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2\right)$$
The fraction goes to f'(b), the second part converges to 3. Plug in b=1, done.

terryds
Staff Emeritus