Sunday, July 10, 2016

Calculate limits limxto+inftyfrac3x1x2+1 and limxtoinftyfrac3x342x2+1



I want to calculate the following limits




lim



In both cases we have indeterminate forms. Using L'Hôpital's rule on \text{(1)} gives



\lim_{x\to+\infty} \frac{3x-1}{x^2+1} = \lim_{x\to+\infty}\frac{3}{2x} = 0



Using L'Hôpital's rule on \text{(2)} gives




\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1} = \lim_{x\to-\infty}\frac{9x^2}{4x} = \lim_{x\to-\infty}\frac{18x}{4} = -\infty



Is this correct?


Answer



In the first case, once you no longer have an indeterminate form, L'Hospital is not needed.



So... After one application of LH, you should obtain \lim \limits_{x\to+\infty}\dfrac 3{2x}, which has the "form" \dfrac 3{\infty} and not \dfrac 3 0, and in any case, is not an indeterminate form, so L'Hospital is no longer warranted. What we do have is the following (after one meager application of LH):



\lim \limits_{x\to+\infty}\dfrac 3{2x} = 0







Your second limit is correct.


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