Question 987267
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*[tex \Large -5\ \not{=}\ \frac{1}{2}] so that one is straight-forward, despite the messy fractions.
 

The other one is straight-forward as well because for *[tex \Large \lim_{x\right{c}}\,f(x)] to exist and be equal to *[tex \Large L], there is no requirement for *[tex \Large f(c)\ =\ L] or to even be defined.


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

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