I'm a student studying maths, I'm taking a real analysis course and I'm looking through some old exams questions. I've arrived at a number of questions that ask me to take the limit of a number of piece-wise functions as $x \rightarrow 2$. Here is the first one.
$$f(x)= \begin{cases} \frac{x^2-4}{x-2} & \text{if }x \neq 2 \\ 6 & \text{if }x = 2 \\ \end{cases} $$
I know if I do some algebra we get...
$$\frac{x^2-4}{x-2} = \frac{(x+2)(x-2)}{x-2}= x+2$$
So we get.
$$f(x)= \begin{cases} x+2 & \text{if }x \neq 2 \\ 6 & \text{if }x = 2 \\ \end{cases} $$
so if we take the limit $$\lim_{x\to2}f(x) = ???$$
I'm not sure weather the answer is 4 or 6. They both seem intuitively correct so perhaps the answer might even be neither. I'm very interested to know!
Thanks for your time!
If you're not sure work the definition of limit. Let $\epsilon>0$.So if $\delta=\epsilon$, for every $x\in \mathbb{R}$ with $0<|x- 2|< \delta$ you have $$|f(x)-4|=|x-2|<\epsilon.$$ So you have that $$\lim_{x\to 2}f(x)=4.$$