Limit of Function

Question

If $\lim_{x\rightarrow 2} \frac {f(x)-5}{x-2}=3$, find $\lim_{x\rightarrow 2} f(x)$.

This is what I have so far.

$$\lim_{x\rightarrow 2} \frac {f(x)-5}{x-2} = \lim_{x\rightarrow 2}\left(f(x) \cdot \frac{1}{x-2}\right)-\frac{5}{x-2}$$

Is this right? What do I do next?

Answer 1

I wouldn't split it up that way, as $\frac{5}{x - 2}$ doesn't approach anything finite as $x \to 2$, so it's not going to help you apply algebra of limits.

Instead, try expressing:

$$f(x) = \frac{f(x) - 5}{x - 2} \cdot (x - 2) + 5$$

and applying algebra of limits. We get

$$f(x) \to 3 \cdot 0 + 5 = 5.$$

Answer 2

I would argue this way: the denominator is going to zero. For the overall initial limit to be finite, what would the numerator have to approach? And what does that say about the desired limit?