Two statements about limits

I think they're both correct. I'm no master with limits either. They are an awkward concept to grasp, but you should take time and fight the repulsion, because limits are the foundation of calculus. You'll be able to solve run of the mill problems without them, but in order to really understand what's going on, you've got to try to understand them.

That said, I didn't really understand them until the second half of calc 2 heh.

 

Both of your statements are false.

*Statement #1 counterexample:
[tex] \text{Let} \; f ( x ) = \frac{{x^2 + x - 6}}{{x - 2}} [/tex]
Then,
[tex] \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \frac{{x^2 + x - 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} x + 3 = 5 [/tex]

but the removable discontinuity
[tex] f\left( 2 \right){\text{ does not exist}}{\text{.}} [/tex]

*Statement #2 counterexample:
[tex] {\text{Let }}f\left( x \right) = 2x - 1\;{\text{and }}g\left( x \right) = x^2 [/tex]

Which satisfies [itex] \forall x \ne 1,\;f\left( x \right) < g\left( x \right) [/itex]

But,
[tex] \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 1 [/tex]

and [itex] 1 =1 [/itex]