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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??
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@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
wait so are we still doing finding the limit by p(a)?
yeah, after cancelling the factor we substitute the value x=2
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