4

It is known that $$\lim_{x \to 0}\frac{f(x)}{x} = -\frac12$$

Solve $$\lim_{x \to 1}\frac{f(x^3-1)}{x-1}.$$

Beforehand, I know that I should aim to get rid of the denominator $(x-1)$ and as such I factor the numerator to get:

$$\lim_{x \to 1}{f(x^2+x+1)}{}.$$

Now that I factored the denominator out, I believe I can insert the 1 in to the limit and I would end up with $f(3)$. Here is where I am confused, how can I incorporate the $-\frac12$ in to this? I figured that since one is approaching $1$ and the other is approaching $0$ there is more to this problem. My guess is that I can simply multiply the two limits to get the answer of $-3/2$.

Since the original limit is simply $f(x) / x$ , all I have to do is multiply it by $x$(in this case it is 3) to get $f(x)$ again.

Am I on the right path?

user382540
  • 352
  • 2
    Since none of the answers mention it - they just show how to do it correctly - I would like to point out the error you made. In general, $$\frac{f((x-1)(x^2+x+1))}{x-1} \ne f(x^2+x+1)$$You don't know how $f$ manipulates its argument, so you cannot cancel something inside of $f$ with something outside. – Paul Sinclair Oct 31 '16 at 01:25

3 Answers3

10

$$\begin{align}\lim_{x \to 1}\frac{f(x^3-1)}{x-1} &= \lim_{x \to 1}\frac{f(x^3-1)}{x-1}\cdot\frac{x^2+x+1}{x^2+x+1} \\ &= \lim_{x\to 1}\frac{f(x^3-1)}{x^3-1}(x^2+x+1) \\ &= \lim_{x\to 1}\frac{f(x^3-1)}{x^3-1}\lim_{x\to 1}(x^2+x+1) \\ &= \lim_{u\to 0}\frac{f(u)}{u}(3) \\ &= -\frac 32\end{align}$$

  • So, what you did was multiply $x-1$ to get it to $x^3 - 1$ to be able to substitute it for the variable u? I'm assuming $u = x-1$? Please correct me wherever possible. I learn by being able to verbally express what's going on in simple terms. – user382540 Oct 30 '16 at 20:00
  • 1
    Actually $u=x^3-1$. But yeah. I knew that we needed to use $\lim \frac{f(u)}{u}$, so I multiplied the denominator by $x^2+x+1$ (but of course to not change the limit, I had to multiply the numerator by that same factor as well) to get it into that form. –  Oct 30 '16 at 20:01
  • How did you get from lim x->1 to lim u->0? – DeadMG Oct 30 '16 at 23:51
  • @DeadMG Change of variables: $u=x^3-1 \implies$ $u\to 0$ as $x\to 1$. –  Oct 31 '16 at 01:11
  • Hey, was coming back to this problem and I'm having troubles explaining to myself why $f(u)/u$ is the same as $f(x)/x$ as they both approach 0. I can solve problems like this, but I just can't verbally explain why this is okay. – user382540 Nov 03 '16 at 16:55
  • 1
    @user382540 Consider $\lim\limits_{x\to 1}\frac{f(x^3-1)}{x^3-1}$. Do the change of coordinates $u=x^3-1$. Then clearly $\frac{f(x^3-1)}{x^3-1} = \frac{f(u)}{u}$, but how does the limit change? Well clearly as $x\to 1$, we have $u = x^3-1 \to 0$. So then $$\lim_{x\to 1}\frac{f(x^3-1)}{x^3-1} = \lim_{u\to 0}\frac{f(u)}{u}$$ –  Nov 08 '16 at 03:29
3

We have

$$\frac{f(x^3-1)}{x-1}=$$

$$(x^2+x+1)\frac{f(x^3-1)}{x^3-1}.$$

and when $x\to 1\;$, your limit is $3.\frac{-1}{2}=\frac{-3}{2}$.

hamam_Abdallah
  • 62,951
2

First of all, note that $$\lim_{x \to 1}\frac{f(x^3-1)}{x-1}=\lim_{x \to 1}\left(\dfrac{f(x^3-1)}{x^3-1}\dfrac{x^3-1}{x-1}\right)=\lim_{x \to 1}\dfrac{f(x^3-1)}{x^3-1}\lim_{x \to 1}\dfrac{x^3-1}{x-1}$$ where the last equality holds if both limits exist. Now, use that,

$$\lim_{x \to 1}\dfrac{f(x^3-1)}{x^3-1}=\lim_{x^3 \to 1}\dfrac{f(x^3-1)}{x^3-1}=\lim_{x \to 0}\dfrac{f(x)}{x}=-\dfrac{1}{2}$$ and

$$\lim_{x \to 1}\dfrac{x^3-1}{x-1}=\lim_{x \to 1}(x^2+x+1)=3.$$

mfl
  • 29,399