Question 364043
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This is an *[tex \Large \frac{\infty}{\infty}] indeterminate form.  Use L'Hôpital's rule.


If *[tex \Large \lim_{x\rightarrow c}\ f(x)\ =\ \lim_{x\rightarrow c}\ g(x)\ =\ 0] or *[tex \Large \pm\infty] and *[tex \Large \lim_{x\rightarrow c}\ \frac{f'(x)}{g'(x)}] exists, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow c}\ \frac{f(x)}{g(x)}\ =\ \lim_{x\rightarrow c}\ \frac{f'(x)}{g'(x)}]


So if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{f(x)}{g(x)}\ =\ \frac{4x\ -\ 5}{x\ -\ 6}]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{f'(x)}{g'(x)}\ =\ \frac{4}{1}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow\infty}\ \frac{4}{1}\ =\ 4]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow\infty}\ \frac{4x\ -\ 5}{x\ -\ 6}\ =\ 4]



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