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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Why 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) \[\lim_{x \rightarrow 0}\frac{e^{2x}-1}{\tan(x)}\]
because for b) it doesn't form \(\dfrac{0}{0}\) since cos (0) =1
while for c) when x=0, \(e^{2x}= e^0 =1\) and it forms \(\dfrac{0}{0}\)

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But if we apply the rule in (b) it becomes solvable but if we apply it in (c) it only becomes more complicated
No, 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?
to 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.
Hopefully 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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