Thread: limit question

Is the answer to this problem, equal to or ?

I keep getting which gives me .

If the answer is , I'm probably making a simple mistake somewhere.

Originally Posted by cinder

Is the answer to this problem, equal to or ?

I keep getting which gives me .

If the answer is , 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)
= -1/16 -------------answer.

Yeah, I see my mistake. Thanks.

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.

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.