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.

Printable View

- May 5th 2006, 03:21 PMcinderlimit 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. - May 5th 2006, 04:38 PMticbolQuote:

Originally Posted by**cinder**

= [(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. - May 5th 2006, 04:56 PMcinder
Yeah, I see my mistake. Thanks. :)

- May 6th 2006, 06:58 AMAfterShock
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 AMcinderQuote:

Originally Posted by**AfterShock**