Exercise 6:(a) Suppose lim_>c f(x) = L. Show that for every ML, there exists & > 0 such that f(x)MCfor 0 lx- cl< 6.(b) Conclude that if limcf(x) = L 0, then there is some small neighborhood near c such thatf (x)0 (except possibly at c).(c) [Hard] Suppose limc f(x) L 0. Prove that limc f(x)1 L-.(d) Conclude, using the product rule for limits, that if limg,c f(x) = L, and lim,c g(x) = M # 0,then limgc f(x)/g(x) = L/M.>C

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Asked Oct 13, 2019
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Exercise 6:(a) Suppose lim_>c f(x) = L. Show that for every M
L, there exists & > 0 such that f(x)
M
C
for 0 lx- cl< 6.
(b) Conclude that if limcf(x) = L 0, then there is some small neighborhood near c such that
f (x)0 (except possibly at c).
(c) [Hard] Suppose limc f(x) L 0. Prove that limc f(x)1 L-.
(d) Conclude, using the product rule for limits, that if limg,c f(x) = L, and lim,c g(x) = M # 0,
then limgc f(x)/g(x) = L/M.
>C
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Exercise 6:(a) Suppose lim_>c f(x) = L. Show that for every M L, there exists & > 0 such that f(x) M C for 0 lx- cl< 6. (b) Conclude that if limcf(x) = L 0, then there is some small neighborhood near c such that f (x)0 (except possibly at c). (c) [Hard] Suppose limc f(x) L 0. Prove that limc f(x)1 L-. (d) Conclude, using the product rule for limits, that if limg,c f(x) = L, and lim,c g(x) = M # 0, then limgc f(x)/g(x) = L/M. >C

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Step 1

a.

Proof:

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Suppose lim f (x) = L Then by the definition oflimit, for every E, there exists a 6 such thatf (x)-L<s when 0< x-c< Assume M L, then for every E, there does not exist any 5 such that f(x)-M when 0< x-c< However, there exist some 5 such thatf(x)-M2s when 0< x-c<6. Hence, it is proved that for every M t L, there exist ao such that f (x) *M when గ గగగగ 0x-c<.

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Step 2

b.

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From part a, for every M L there exist a o such that f(x)M when 0 < x -c<6 That is, for lim f (x) = L M, there exist a o neighborhood around c such that f(x)M Take M 0 Hence, it can be concluded that for lim f (x) = L 0, there exist a 5 neighborhood around c such that f (x)#0.

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Step 3

c.

Proof:

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Suppose lim f (x) = L 0 Let S lim f(x) Then, we have S = L 0 1 Take reciprocal on both sides and obtain S L Replace the value of S. Thus, we obtain lim f(x)= L1 x-c -1 Hence, it is proved that limf (x)= L

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