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If lim (x->a) |f(x)| = 0, prove that lim (x->a) f(x) =0?
I can't figure out how to prove this. I know that it involves the squeeze theorem. But the Squeeze Theorem features f(x) <= g(x) <= h(x). How can I get started on this?
2 Answers
- сhееsеr1Lv 71 decade agoFavorite Answer
Given:
lim |f(x)| = 0
you also know:
lim -|f(x)| = - lim |f(x)| = -0 = 0
Now use the fact:
- |f(x)| ≤ f(x) ≤ |f(x)|
To say:
lim -|f(x)| ≤ lim f(x) ≤ lim |f(x)|
The left and right limits are zero, thus so too is the middle:
lim f(x) = 0
Source(s): NOTE: "mathman" has given a nonsense answer. Please do not let it confuse you. There is no guarantee that near the point x=a, f(x) is always positive or always negative, so you cannot substitute |f(x)| = f(x) or anything like that!! Assuming that's what he's suggested, which isn't even very clear. - mathman241Lv 61 decade ago
Note 0 is nether positive or negative
Hence if |f(x)| =0 = -f(x) = f(x)
|A| = A or -A
so
-|A|<= A <= |A|
