Just had an exam where the question was roughly:

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Using an epsilon delta definition of limit, prove that the limit

$\displaystyle \lim_{x\rightarrow 1}x^2 + 3 = 4$

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I am not very confident with proving the limit definition of quadratics..

So I folded and decided to do it by two sided limit.

Rough attempt

For an $\displaystyle \epsilon > 0$, there exists a $\displaystyle \delta > 0$, such that $\displaystyle |x-a|<\delta\Rightarrow |f(x)-L|<\epsilon$:

Left hand limit:

$\displaystyle 1 - \delta < 1 \Rightarrow \delta > 0$

$\displaystyle 4 - \epsilon < 4 \Rightarrow \epsilon > 0$

Right hand limit:

$\displaystyle 1 < 1 + \delta \Rightarrow \delta > 0$

$\displaystyle 4 < 4 + \epsilon \Rightarrow \epsilon > 0$

There exists an $\displaystyle \epsilon > 0$ and a $\displaystyle \delta > 0$ on both sides of 1, which approach the value of 4.

Thefore the limit does exist, and it is 4. ▫

Pretty sure I messed up monumentally.. Linear functions are so much easier to prove than quadratics. :/