Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim
ID: 3139660 • Letter: E
Question
Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.)lim (square root(4 + h) ? 2)/h
h approaches 0
Explanation / Answer
No, do not multiply by conjugate. You would only do so if you have a situation that gives you 0/0 So if limit were lim [h->0] 3(v(3h+1) - 1) / 3h which evaluates to 3(v1-1)/0 = 0/0 then you would multiply by conjugate, giving you lim [h->0] 3 (v(3h+1) - 1) / 3h = lim [h->0] (v(3h+1) - 1) / h = lim[h->0] (v(3h+1)-1) (v(3h+1)+1) / (h (v(3h+1)+1) = lim[h->0] ((3h+1) - 1) / (h (v(3h+1)+1) = lim[h->0] 3h / (h (v(3h+1)+1) = lim[h->0] 3 / (v(3h+1)+1) which is what you started off with! Evaluating directly we get = 3 / (v(0+1)+1) = 3 / (1+1) = 3/2 -------------------- So there was no need to multiply by conjugate, since you did not get 0/0
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