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## anonymous one year ago ques

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1. anonymous

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

2. Loser66

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

3. Loser66

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

4. anonymous

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

5. Loser66

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?

6. Loser66

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.

7. Loser66

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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