Finding multiple variables to make a function continuous

Question

Given the function:

give values for $a$ and $b$ in order that $f(x)$ is continuous at $x=1$

I can't figure out how to calculate $a$. When $x$ is 1 the numerator is 0,. How do I get the limits on either side to be equal?

Answer 1

As $b(x-1)$ is always $0$ at $x=1$, $b$ is arbitrary. Then we must achieve

$$\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}3\frac{(x-1)(x-2)}{1-ax}=\lim_{x\to 1^-}3\frac{1-x}{1-ax}=3$$ and this works with $a=1$ (any other value of $a$ makes $f(1^-)=0$).

Answer 2

Hints: compute $\lim_{x \to 1+0}f(x)$ and $f(1)$. Furthermore compute $\lim_{x \to 1-0}f(x)$ in the cases $a=1$ and $a \ne 1.$

$f$ is continuous at $x=1 \iff \lim_{x \to 1+0}f(x)=f(1)=\lim_{x \to 1-0}f(x).$

Answer 3

OK, got it. f(1)=3 As well as lim→1+0()=3. For the lim→1−0() we assume a=1 which allows us to factor the denominator to -(x-2). The numerator is factored to 3(x-2)(x-1). The (x-2) in the numerator and denominator cancel out which leaves us with 3(x-1)/-1. Solving for x=1 we get 3 which confirms continuity for a=1. If ≠1 we would not be able to factor and would always get 0 in the numerator so a could only be 1. b can be anything because we would always get 3 for f(1) and lim→1+0()