sin(x^2-4)/(x-2) --> 2xcos(x^2-4)/1 --> 4
ln(x-3)/(x-5) --> (1/(x-3))/5 -> 0
(x-pi/4)tan(2x) = (x-pi/4)/cot(2x) --> 1/(-2csc^2(2x)) = -1/2 sin^2(2x) --> -1/2
5) Use the L’Hopital’s method to evaluate the following limits. In each case, indicate what type of limit it is ( 0/0, ∞/∞, or 0∙∞)
lim x→2 sin(x^2−4)/(x−2) =
lim x→+∞ ln(x−3)/(x−5) =
lim x→pi/4 (x−pi/4)tan(2x) =
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