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Mathematics Limits & proof Question?
Hi I need some desperate help on these question any help will be very much appreciated
A.)
Prove each of the following statements.
(i) lim x→1 (2x^3 − 5x^2 + 2x + 1) / (x^2 − 4x + 3) = 1
(ii) lim x→0 (sin x) / e^(sin x) − 1 = 1
(iii) lim x→∞ 4x^4 − log(e)x / 2e^x + 3x^2 = 0
B.)
Use the ε–δ definition of continuity to prove that the function
f(x) = 2x^2 − 3x
is continuous at 2.
C.)
Prove that the function
f(x) = (x − 2) / (x + 1)
is uniformly continuous on the interval [1, 3], stating any results from
the module which you use.
2 Answers
- kbLv 79 years agoFavorite Answer
A) If you need ε–δ proofs, please let me know. (How rigorous do these need to be?)
(i) lim(x→1) (2x^3 − 5x^2 + 2x + 1) / (x^2 − 4x + 3)
= lim(x→1) (2x^2 - 3x - 1)(x - 1) / [(x - 3)(x - 1)]
= lim(x→1) (2x^2 - 3x - 1)/(x - 3)
= (2 - 3 - 1)/(1 - 3), via quotient rule for limits
= 1.
(ii) lim(x→0) (sin x) / (e^(sin x) − 1)
= lim(t→0) t / (e^t − 1), letting t = sin x
= 1 / [lim(t→0) (e^t − 1)/t]
= 1 / [(d/dt) e^t {at t = 0}], via definition of a derivative
= 1/e^0
= 1.
(iii) lim(x→∞) (4x^4 − ln x) / (2e^x + 3x^2)
= lim(x→∞) (4x^4/e^x − ln x/e^x) / (2 + 3x^2/e^x), dividing each term by e^x
= (0 − 0) / (2 + 0), by L'Hopital's Rule or otherwise.
----------------------
B) Note that |f(x) - f(2)|
= |2x^2 - 3x - 2|
= |(2x + 1)(x - 2)|
= |2((x - 2) + 2) + 1| * |x - 2|
= |2(x - 2) + 5| |x - 2|
≤ 2 |x - 2|^2 + 5 |x - 2|
< 2 |x - 2| + 5 |x - 2|, assuming that |x - 2| < 1
= 7|x - 2|.
So given ε > 0, let δ = min{1, ε/7}. Then, |x - 2| < δ implies that
|f(x) - f(2)| < 7|x - 2| < 7(ε/7) = ε.
Hence, f is continuous at x = 2.
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C) The short version is that f is continuous (quotient of continuous functions, and the denominator is nonzero) on a compact set.
An ε–δ is like this:
Given ε > 0, we need δ > 0 such that for all x, y in [1, 3] with |x - y| < δ, we have |f(x) - f(y)| < ε.
To this end, note that since f(x) = (x - 2)/(x + 1) = 1 - 3/(x+1),
|f(x) - f(y)|
= |-3/(x+1) - (-3/(y+1))|
= 3|1/(y+1) - 1/(x+1)|
= 3 |x - y| / ((x+1)(y+1)), by combining the fractions (and noting x, y > 0)
≤ 3 |x - y| / ((1+1) (1+1)), since 1/(t+1) is decreasing on [1,3]
= (3/4) |x - y|.
So given ε > 0, let δ = min{1, 4ε/3}. Then, for all x, y in [1, 3] with |x - y| < δ, we have
|f(x) - f(y)| ≤ (3/4)|x - y| ≤ (3/4)(4ε/3) = ε.
Hence, f is uniformly continuous on [1,3].
I hope this helps!
