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A positive number $\epsilon$ and the limit $L$ of a function $f$ at $-\infty$ are given. Find a negative number $N$ such that $|f(x)-L|<\epsilon$ if $x<N$.$$\lim _{x \rightarrow-\infty} \frac{1}{x^{2}}=0 ; \epsilon=0.01$$

$|x|>10$

Calculus 1 / AB

Chapter 1

LIMITS AND CONTINUITY

Section 4

Limits (Discussed More Rigorously)

Functions

Limits

Continuous Functions

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a problem. Number 46. Assume X is negative. Ah, went over Excess words upset you. It's one of that. All points or one so one over X. It's weird. It's more than opening over what X is. Ah, it's is negative, and X squared is positive. So from here we can say that there is a broken for that. So it's square is bigger than one over open toe ones for excess bread. Bigger that one were open. One of these 100 so X is a smarter than negative square or 100 where it's is negative. As we sit here, X is smaller than negative then so N is they get the there.

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