Question

ques

Answer 1

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)
\[\lim_{x \rightarrow 0}\frac{e^{2x}-1}{\tan(x)}\]

Answer 2

because for b) it doesn't form \(\dfrac{0}{0}\) since cos (0) =1

Answer 3

while for c) when x=0, \(e^{2x}= e^0 =1\) and it forms \(\dfrac{0}{0}\)

Answer 4

But if we apply the rule in (b) it becomes solvable but if we apply it in (c) it only becomes more complicated

Answer 5

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?

Answer 6

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.

Answer 7

Hopefully I was wrong on the last sentence so that I am happy to help you out if you have other questions later.