anonymous
  • anonymous
ques
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • 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) \[\lim_{x \rightarrow 0}\frac{e^{2x}-1}{\tan(x)}\]
Loser66
  • Loser66
because for b) it doesn't form \(\dfrac{0}{0}\) since cos (0) =1
Loser66
  • Loser66
while for c) when x=0, \(e^{2x}= e^0 =1\) and it forms \(\dfrac{0}{0}\)

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anonymous
  • anonymous
But if we apply the rule in (b) it becomes solvable but if we apply it in (c) it only becomes more complicated
Loser66
  • 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?
Loser66
  • 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.
Loser66
  • 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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