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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why is L'Hopital's rule not applicable for (b) but it is applicable for (c), when I think it should be the other way around (b) \[\lim_{x \rightarrow 0}\frac{\cos(x)}{x}\] (c) \[\lim_{x \rightarrow 0}\frac{e^{2x}1}{\tan(x)}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1because for b) it doesn't form \(\dfrac{0}{0}\) since cos (0) =1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1while for c) when x=0, \(e^{2x}= e^0 =1\) and it forms \(\dfrac{0}{0}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But if we apply the rule in (b) it becomes solvable but if we apply it in (c) it only becomes more complicated

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1No, no, no!!! you are allowed to apply L'Hospital rule if and only if it forms the special cases like (0/0), (0^ infinitive).... if the expression is not one of the special case, you can't. To b) it is just \(=\infty\), why is it complicated?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1to c) it is not at all \((e^{2x}+1)'= 2e^{2x}\) and \((tan(x))'= sec^2(x)=1/cos^2(x)\) when x =0, the limit of numerator =2, the limit of denominator =1 and you get 2 as the final result. how complicated it is UNLESS you try to get the free answer from me.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hopefully I was wrong on the last sentence so that I am happy to help you out if you have other questions later.
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