Need to find the Limit If $\lim_{x\to 1}\frac{f(x)-2}{x-1}=3$, where $f$ is continuous over $\mathbb{R}$.
Find $\lim_{x\to 1}\frac{x^3f(x)-f(1)}{x-1}$.
I tried a lot to solve this question but I didn't succeed. Could you please give me a hint.
 A: Given $\lim_{x\to 1}\frac{f(x)-2}{x-1}=3$ and $f$ is continuous:
$$\lim_{x\to 1}\frac{x^3f(x)-f(1)}{x-1}=\\
\lim_{x\to 1}\frac{(x^3f(x)-2x^3)+(2x^3-2x^2)+(2x^2-2x)+(2x-2)+(2-f(1))}{x-1}=\\
\lim_{x\to 1} x^3\cdot \lim_{x\to 1}\frac{f(x)-2}{x-1}+\lim_{x\to 1} 2x^2+\lim_{x\to 1} 2x+2+0=\\
3+2+2+2+0=9.$$
A: The first statement means $f(1)=2$ and $f'(1)=3$. Then
$$f(1+h)=2+3h+o(h)$$ and so
$$\lim_{x\to1}\frac{x^3 f(x)-f(1)}{x-1}
=\lim_{h\to0}\frac{(1+h)^3(2+3h+o(h))-2}{h}$$
etc.
A: $$\lim_{x\to 1}\frac{f(x)-2}{x-1}=3 \implies f(x)=(x-1)g(x)+2$$
Where $g(x)$ is a continuous function and $$\lim _{x\to1} g(x)=3$$
Thus $$\lim_{x\to 1}\frac{x^3f(x)-f(1)}{x-1}=   \lim_{x\to 1}\frac{x^3(x-1)g(x)+2x^3-f(1)}{x-1}$$
$$=g(1) +\lim_{x\to1}\frac{2x^3-2}{x-1}=3 +6=9$$
A: Thanks for your hints, but I think the following is easier for an undergraduate student 
The first statement means : $f(1) = 2$ and $f'(1) = 3$, then 
$\lim_{x \to 1}\frac{x^3f(x)-f(1)}{x-1} = \lim_{x \to 1}\frac{x^3f(x)-2}{x-1}$
If we use the long division, we will get 
$\lim_{x \to 1}\frac{x^3f(x)-2}{x-1} = \lim_{x \to 1}f(x)(x^2+x+1)+\frac{f(x)-2}{x-1} = \lim_{x \to 1}f(x)(x^2+x+1)+\lim_{x \to 1}\frac{f(x)-2}{x-1} = f(1).3 + 3 = 6+3=9$
Or, we can use L'hopitals rule 
$\lim_{x \to 1}\frac{x^3f(x)-2}{x-1} = \lim_{x \to 1}\frac{3x^2f(x) + x^3f'(x)}{1} = 3f(1) + f'(1) = 6+3=9$
A: Second part:
$\dfrac {[(x-1)+1]^3f(x)-f(1)}{x-1}=$
${\tiny \dfrac {[(x-1)^3+3(x-1)^2+3(x-1)^1+ 1]f(x)-f(1)}{x-1}=}$
${\tiny \dfrac{(x-1)[3(x-1)^2+3(x-1)^1 +3] f(x)+f(x)-f(1)}{x-1}=}$
$[3(x-1)^2+3(x-1)^1 +3]f(x) +\dfrac{f(x)-f(1)}{x-1};$
Take the limit $x \rightarrow 1$ making use of the continuity of $f$ , of $f(1)=2$, and of $\lim_{ x \rightarrow 1}\dfrac{f(x)-f(1)}{x-1}=3$.
A: Suppose that $f(x)=f(1)+f'(1)(x-1)+o((x-1)^2)$.
From the condition, we have $f(1)=2$ and $f'(1)=3$.
The result is $9$.
