Question 979519
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right\infty}\,\frac{4x^2\ -\ 3x\ + 1}{4x^2\ +\ 6x\ -\ 5}]


This is a L'Hôpital indeterminate form *[tex \Large \frac{\infty}{\infty}]. So *[tex \Large \lim\,\frac{f(x)}{g(x)}\ =\ \lim\,\frac{f'(x)}{g'(x)}].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right\infty}\,\frac{8x\ -\ 3}{8x\ +\ 6}]


This is still the indeterminate form *[tex \Large \frac{\infty}{\infty}].  So apply L'Hôpital again.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right\infty}\,\frac{8}{8}\ =\ 1]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \