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  • one year ago

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  1. anonymous
    • one year ago
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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) \[\lim_{x \rightarrow 0}\frac{e^{2x}-1}{\tan(x)}\]

  2. Loser66
    • one year ago
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    because for b) it doesn't form \(\dfrac{0}{0}\) since cos (0) =1

  3. Loser66
    • one year ago
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    while for c) when x=0, \(e^{2x}= e^0 =1\) and it forms \(\dfrac{0}{0}\)

  4. anonymous
    • one year ago
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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

  5. Loser66
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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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