0 if 0 <|x − 0| &lt

Exercise 2.4 P.117

26. Prove the statement using the ε and δ definition of a limit.

lim

x→0x³= 0

< pf >

Given ε > 0 take δ = ε¹³ > 0 if 0 <|x − 0| < δ = ε¹³ then |x³− 0| = |x|³< δ³= ε

30. Prove the statement using the ε and δ definition of a limit.

lim

x→2(x²+ 2x− 7) = 1

< pf >

Given ε > 0 take δ = min{1,^ε₇} > 0 if 0 <|x − 2| < δ

then |(x²+ 2x− 7) − 1| = |x²+ 2x− 8| = |x + 4||x − 2| < |x + 4|δ < 7δ < ε (∵ |x − 2| < δ ≤ 1 ∴ |x + 4| < 7)

P.118

36. Prove that

lim

x→2

1 x =1

2

< pf >

Given ε > 0 take δ = min{1, 2ε} > 0 if 0 <|x − 2| < δ

then |(¹_x)−¹₂| = |²_2x^−x| < |_2x^δ | < |^δ₂| < ε (∵ |x − 2| < δ ≤ 1 ∴ 1 ≤ x ≤ 3 ∧ |2x| > 2) 37. Prove that

lim

x→a

√x =√ a

< pf >

Given ε > 0 take δ =√ aε > 0 if 0 <|x − a| < δ

then |√ x−√

a| = |^√x_x+^−a√_a| < ^√_x+^δ√_a <|^√δ_a| < ε 39. If the function f is defined by

f (x) =

{ 0 , if x is irrational 1 , if x is rational prove that lim

x→0f (x) does not exist.

< pf >

Take ε = ¹₂ ∀δ > 0 if 0 < |x − 0| < δ

∵ |f(x) − f(0)| =

{ |0 − 0| = 0 , if x is irrational

|1 − 0| = 1 , if x is rational

∴ lim

x→0f (x) does not exist 42. Prove, using Definition 6, that

lim

x→−3

1

(x + 3)⁴ =∞

< pf >

Given M > 0 take δ = √4¹

M > 0 if 0 <|x − (−3)| < δ ⇒ |x + 3| < δ then f (x) = _(x+3)¹ ₄ =_|x+3|¹ ₄ > _δ¹₄ = M

1

43. Prove that

lim

x→0⁺ln x =−∞

< pf >

Given N < 0 take δ = e^N > 0 if 0 < x− 0 < δ ⇒ x < δ then f (x) = ln x < ln δ = N 44. Suppose that lim

x→af (x) =∞ and lim

x→ag(x) = c, where c is a real number.

Prove each statement.

(a) lim

x→a[f (x) + g(x)] =∞ (b) lim

x→a[f (x)g(x)] =∞ if c > 0 (c) lim

x→a[f (x)g(x)] =−∞ if c < 0

< pf >

∵ lim

x→af (x) =∞ ∧ lim

x→ag(x) = c

∴Given M1, ε > 0∃ δ1, δ2 then take δ = min{δ1, δ2} if 0 <|x − a| < δ

then f (x) > M1 ∧ |g(x) − c| < ε (i.e c − ε < g(x) < c + ε) (a) Now for any M > 0 take M1= M− c + ε

if 0 <|x − a| < δ

then f (x) + g(x) > M1+ c− ε = M

(b) Now for any M > 0 take M1= _c^M_−ε ∧ c − ε > 0 if 0 <|x − a| < δ

then f (x)g(x) > M₁(c− ε) = M

(c) Now for any N < 0 take M₁=_c+ε^N ∧ c + ε < 0 if 0 <|x − a| < δ

then f (x)g(x) < M₁(c + ε) = N

2