Limits Calculus

Question

Limit of function (m³-8)/(x-2) as m approaches 2.

How do you go about finding limits to these types of problems where the function has two different variables?

Answer 1

As Mark M. suggested above it's probably a typo and the limit should be lim_m→2 [(m³– 8) / (m – 2)].

Since it's a 0/0 case I used L'Hôpital's rule, taking the derivative of numerator and denominator and calculated the resulting limit.

d/dm(m³– 8) = 3m² and d/dm(m – 2) = 1

So lim_m→2 [(m³– 8) / (m – 2)] = lim_m→2 [3m² / 1] = 3(2)² = 12

Answer 2

Probably a typo. If "x" is supposed to be "m", then we have

lim_m→2 [(m³-8) / (m-2)] = lim_m→2 [ (m-2)(m² + 2m + 4) / (m-2) ] = lim_m→2 [ m² + 2m + 4 ] = 12

Answer 3

The limit as m---> 2 is (2^3 - 8)/x-2) = 0/(x-2) = 0