# limit question

• May 5th 2006, 03:21 PM
cinder
limit question
Is the answer to this problem, $\lim_{x \to 4}\frac{\frac{1}{m}-\frac{1}{4}}{m-4}$ equal to $\frac{-1}{16}$ or $\frac{1}{16}$?

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

If the answer is $\frac{-1}{16}$, I'm probably making a simple mistake somewhere.
• May 5th 2006, 04:38 PM
ticbol
Quote:

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

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

If the answer is $\frac{-1}{16}$, 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)
• May 5th 2006, 04:56 PM
cinder
Yeah, I see my mistake. Thanks. :)
• May 6th 2006, 06:58 AM
AfterShock
Since you get the limit of 0/0 (an indeterminate form) when you substitute 4 in for x, this problem can also be solved using l'Hopital's rule.
• May 6th 2006, 10:30 AM
cinder
Quote:

Originally Posted by AfterShock
Since you get the limit of 0/0 (an indeterminate form) when you substitute 4 in for x, this problem can also be solved using l'Hopital's rule.

Yes, but my calculus teacher won't give us credit if we do it that way. ;)