Paradoxica
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lol he didn't mention it that was me inserting my commentary into his quote.If you need to estimate the integrand anyway, I would just go ahead and use the Taylor bounds you mentioned.
=\frac{1}{x}\int_x^{4x} \cos(1/u)\, du\Rightarrow \\ \\ 3=\frac{1}{x}\int_x^{4x}\, du\geq F(x)\geq \frac{1}{x}\int_x^{4x}(1-\frac{1}{2u^2})\, du = 3-\frac{3}{8x^2}\\ \\ \Rightarrow \lim_{x\rightarrow\infty} F(x)=3. )
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Conditions on L'Hopital's rule (2 Viewers)
- Thread starter leehuan
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seanieg89
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Lol, it always catches me off guard when people edit the things they quote.
0/0, since numerator goes to ln(1) = 0.I was basically prepared to bash this with L'Hopital's rule just like the answers did, but what "indeterminate form" is this supposed to be again?
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