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# If $f"$ is continuous, show that$$\displaystyle \lim_{h\to 0} \frac{f(x + h) - 2f(x) + f(x - h)}{h^2} = f"(x)$$

## $\lim _{x \rightarrow 0}\left(2 \cos 2 x+3 a x^{2}+b\right)=0 \Rightarrow$$2 \cos (2 \cdot 0)+3 a \cdot 0^{2}+b=0 \Rightarrow 2+b=0 \Rightarrow b=-2$

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we know that we can rewrite this limit as H A purchase. Zero. That's a crime. Expose Age miss off prime X minus age times Negative one, which is of prime of acts, cause extra zeros. Just acts plus f from of acts because X minus zero is also just acts divide by two, which gives us off prime of acts because two F products over two is just one F part of X. Okay, now we know that we can differentiate with respect to H instead of acts we have B is negative, too.

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