Reviewing Calculus, I am facing the problem:


$$f(x)= \begin{cases} \sqrt[3]{x^2-8x^3}+ax+b,& \text{if } x\in\mathbb Q\\\\x\sin\big(\frac{1}{x}\big),& x\in \mathbb R-\mathbb Q \end{cases}$$ has a limit at $ +\infty$, what would $ab$ be?

I doubted if I could treat this function as other piecewise function with some known domains (like $-7<x \leq 3$) or not? When I referred to the answer, it gave me another approach which admitted my guess and is my question.

Let the functions $f_1(x)$ and $f_2(x)$ have limits on $\mathbb R$ when $x\to +\infty$ so the function:

$$f(x)= \begin{cases} f_1(x),& x\in\mathbb Q\\\\f_2(x),& x\in \mathbb R-\mathbb Q \end{cases}$$ has limit at $+\infty$ if $\lim_{x\to +\infty}f_1(x)=\lim_{x\to +\infty}f_2(x)$

May I ask someone explain this hint? Thanks.


It's easy to compute that

$$\lim_{x\to\infty} x\sin \frac 1x = \lim_{x\to\infty} \frac{\sin \frac1x}{\frac 1x} = \lim_{t\to 0^+} \frac{\sin t}t = 1. \tag{1}$$

So for $\lim_{x\to \infty} f(x)$ to exist we must have that

$$\lim_{x\to\infty} \sqrt[3]{-8x^3+x^2} + ax + b = 1. \tag{2} $$

You'll see that you'll have to pick $a$ such that the limit in $(2)$ even exists and $b$ such that it has the right value. Think about what happens if the limit in $(2)$ exists but doesn't equal $1$. Can you see why $\lim_{x\to\infty}f(x)$ doesn't exist then?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.