If u wanna take lim x- 2 ([x^2+4x-12]/[x^2-2x]) .. why can you sub x=2 in? isnt the function not continuous at x=2? lim x-a = p(a) only if function is continuous no??

Question

If u wanna take lim x-> 2 ([x^2+4x-12]/[x^2-2x]) .. why can you sub x=2 in? isnt the function not continuous at x=2? lim x->a = p(a) only if function is continuous no??

ash2326 · Answer

@powerangers69 There are many types of discontinuities, this is a removable type of discontinuity. We cancel the factor x-2 from numerator and denominator and then we are able to evaluate this limit

anonymous · Answer

wait so are we still doing finding the limit by p(a)?

ash2326 · Answer

yeah, after cancelling the factor we substitute the value x=2