Need to find the Limit

Question

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.

Answer 1

Given $\lim_{x\to 1}\frac{f(x)-2}{x-1}=3$ and $f$ is continuous: $$\lim_{x\to 1}\frac{x^3f(x)-f(1)}{x-1}=\\ \lim_{x\to 1}\frac{(x^3f(x)-2x^3)+(2x^3-2x^2)+(2x^2-2x)+(2x-2)+(2-f(1))}{x-1}=\\ \lim_{x\to 1} x^3\cdot \lim_{x\to 1}\frac{f(x)-2}{x-1}+\lim_{x\to 1} 2x^2+\lim_{x\to 1} 2x+2+0=\\ 3+2+2+2+0=9.$$

Answer 2

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 $9$.

Answer 3

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.

Answer 4

$$\lim_{x\to 1}\frac{f(x)-2}{x-1}=3 \implies f(x)=(x-1)g(x)+2$$

Where $g(x)$ is a continuous function and $$\lim _{x\to1} g(x)=3$$

Thus $$\lim_{x\to 1}\frac{x^3f(x)-f(1)}{x-1}= \lim_{x\to 1}\frac{x^3(x-1)g(x)+2x^3-f(1)}{x-1}$$

$$=g(1) +\lim_{x\to1}\frac{2x^3-2}{x-1}=3 +6=9$$

Answer 5

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 $\lim_{x \to 1}\frac{x^3f(x)-2}{x-1} = \lim_{x \to 1}\frac{3x^2f(x) + x^3f'(x)}{1} = 3f(1) + f'(1) = 6+3=9$

Answer 6

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 $\lim_{ x \rightarrow 1}\dfrac{f(x)-f(1)}{x-1}=3$.