# evaluate limit using squeeze theorem

• June 7th 2012, 06:55 AM
sluggerbroth
evaluate limit using squeeze theorem
lim f(x) as x goes to 2 where 2x-1 (< or equal) f(x) (< or equal) x^2-2x+3, x does not = 2

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• June 7th 2012, 07:01 AM
Reckoner
Re: evaluate limit using squeeze theorem
Quote:

Originally Posted by sluggerbroth
lim f(x) as x goes to 2 where 2x-1 (< or equal) f(x) (< or equal) x^2-2x+3, x does not = 2

We have $2x-1\leq f(x)\leq x^2-2x+3$ for all $x\neq2$.

$\lim_{x\to2}\left(2x-1\right) = 3$ and $\lim_{x\to2}\left(x^2-2x+3\right) = 3$. So, by the Squeeze Theorem, $\lim_{x\to2}f(x)=\mathrm?$