- Thread starter dome123
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SEE HERE one needs to Pro Premium edition to get the steps.

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Hi,

I have a problem which needs to be solved without using L'Hosiptal rule.

Can anyone help me?

Thanks.View attachment 15947

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Hint:

If you factor out \(\displaystyle \sqrt[3]{cos(x)}\) from the numerator - what would you get?

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No, it isn't too hard. Computers are not always smarter than people.Well, pka thanks for your reply , I know it's not easy but this is the question on my exam and I was struggling to solve it but now I see it is too hard.

Thanks anyway

The suggests that have been given allowed me to work it out in about two square inches of paper. (Well, I write very small -- that accounts for some of my errors.) Your substitution t = cos(x) does make things a lot easier to write, factoring out t^(1/3) helps more, and some conjugates and the like finish the job. Don't give up!

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You can make any substitution you like. How did this one go? Remember if cos(x)=t, then sin^2(x)= 1 - t^2Is it possible to do a substitution where cosx=t?

\(\displaystyle cos(x) = t^6 \implies \sqrt{cos(x)} = t^3,\ \sqrt[3]{cos(x)} = t^2,\ cos^2(x) = t^{12},\)

\(\displaystyle sin^2(x) = 1 - cos^2(x) = 1 - t^{12} = (1 - t^3)(1 + t^3)(1 + t^6) = (1 - t)(1 + t + t^2)(1 + t^3)(1 + t^6) \text { and}\)

\(\displaystyle t^3 - t^2 = t^2((t = 1) = -\ t^2(1 - t).\)

\(\displaystyle \therefore y = \dfrac{\sqrt{cos(x)} - \sqrt[3]{cos(x)}}{sin^2(x)} = \dfrac{t^3 - t^2}{1 - t^{12}} =\)

\(\displaystyle \dfrac{-\ t^2 \cancel {(1 - t)}}{\cancel {(1 - t)}(1 + t + t^2)(1 + t^3)(1 + t^6)} = -\ \dfrac{t^2}{(1 + t + t^2)(1 + t^3)(1 + t^6)}.\)

\(\displaystyle \lim_{x \rightarrow 0} cos(x) = 1 \implies \lim_{x \rightarrow 0} t^6 = 1 \implies \lim_{x \rightarrow 0} t^n =1 \implies\)

\(\displaystyle \lim_{x \rightarrow 0} y = -\ \dfrac{1^2}{(1 + 1^1 + 1^2)(1 + 1^3)(1 + 1^6)} = -\ \dfrac{1}{3(2)(2)} = -\ \dfrac{1}{12}.\)

If you ask a machine just to give you the answer, the reason for the answer will not be apparent.