# Thread: limit question

1. ## limit question

Is the answer to this problem, $\displaystyle \lim_{x \to 4}\frac{\frac{1}{m}-\frac{1}{4}}{m-4}$ equal to $\displaystyle \frac{-1}{16}$ or $\displaystyle \frac{1}{16}$?

I keep getting $\displaystyle \lim_{x \to 4}(\frac{m-4}{4m})(\frac{1}{m-4})$ which gives me $\displaystyle \frac{1}{16}$.

If the answer is $\displaystyle \frac{-1}{16}$, I'm probably making a simple mistake somewhere.

2. Originally Posted by cinder
Is the answer to this problem, $\displaystyle \lim_{x \to 4}\frac{\frac{1}{m}-\frac{1}{4}}{m-4}$ equal to $\displaystyle \frac{-1}{16}$ or $\displaystyle \frac{1}{16}$?

I keep getting $\displaystyle \lim_{x \to 4}(\frac{m-4}{4m})(\frac{1}{m-4})$ which gives me $\displaystyle \frac{1}{16}$.

If the answer is $\displaystyle \frac{-1}{16}$, I'm probably making a simple mistake somewhere.
See if you can simplify the (1/m -1/4)/(m-4).

= [(4 -m)/4m] / (m-4)
See those (4-m) and (m-4), they are the same but one is the negative of the other. If they are exactly the same, then they will cancel each other. So we make them exactly the same.
Say, we turn (4-m) into (m-4). The step is one of the simplest switch in Math. (4-m) becomes -(m-4). One way it's done is by multiplying the (4-m) by (-1 / -1), which is actually 1.
(4-m)*(-1 / -1) = (-4 +m) / -1 = -(-4 +m) = -(m-4). Easy.

So,
[(4-m)/4m] / (m-4)
= [-(m-4)/4m] / (m-4)
= -1/(4m)

Therefore,
Limit(m-->4) [(1/m -1/4)/(m-4)]
= Limit(m->4) [-1/(4m)]
= -1/(4*4)