LIMIT AND CONTINUITY

Solutions:

1. For every ϵ > 0 , there exists δ > 0 such that

| f(x) − l |< ϵ for all x with 0 <| x − a |< δ. (or For every ϵ > 0 , there exists a corresponding δ > 0 such that for all x , 0 <| x − a |< δ ⇒ | f(x) − l |< ϵ. )

2. Arbitrarily give ϵ > 0.

lim_{x→a} f(x) = l

⇒ there exists δ_{1} > 0 such that (0 <| x − a |< δ_{1} ⇒| f(x) − l |< \frac{ϵ}{2} ) … (1)

lim_{x→a} g(x) = m

⇒ there exists δ_{2} > 0 such that

(0 <| x − a |< δ_{2} ⇒| g(x) − m |< \frac{ϵ}{2} ) … (2) Take δ = min{δ_{1}, δ_{2}} , then

0 <| x − a |< δ

⇒ \{ \begin{matrix}0 <| x − a |< δ_{1} \\ 0 <| x − a |< δ_{2} \end{matrix}

⇒ \{ \begin{matrix}| f(x) − l |< \frac{ϵ}{2}  ( ^{·}_{·}^{·}(1) ) \\ | g(x) − m|< \frac{ϵ}{2}  ( ^{·}_{·}^{·}(2) ) \end{matrix}

⇒| (f(x) + g(x) − (l + m) | =| (f(x) − l) + (g(x) − m) | ≤| f(x) − l | + | g(x) − m | < ϵ ·^{·}_{·} For an arbitrarily given ϵ > 0 , there exist δ > 0 such that

0 <| x − a |< δ_{1} ⇒| f(x) + g(x) − (l + m) |< ϵ.

Hence lim_{x→a} f (x) + g(x) = l + m.

3. Suppose g(x) ≤ f(x) ≤ h(x) for all x in some neighborhood of a , and x ≠ a. If lim_{x→a} g(x) = l and lim_{x→a} h(x) = l, then lim_{x→a} f(x) = l.

4. lim_{x→a} f(x) exists, and lim_{x→a} f(x) = f(a)

5. (1) 3 (2) \frac{1}{2} (3) does not exist

6. (i) (1)1 (2) 2 (3) does not exists

(ii) g is not continuous at 0.

·_{·}^{·} lim_{x→0} g(x) does not exist.

7. (i) (1) 2 (2) 2 (3) 2

(ii) h is continuous at -1.

·_{·}^{·} lim_{x→−1} h(x) = h(−1)

8. (i) (1) 0 (2) does not exist (ii) 0

9. lim_{x→0} sin \frac{1}{x} does not exist.

·_{·}^{·} When x approaches 0 , the values of sin \frac{1}{x} change between -1 and 1 ; they do not approach a fixed number.

10. − | x |≤ x sin(\frac{1}{x}) ≤| x | for all x ≠ 0, and

\{ \begin{matrix}lim_{x→o} | x |= 0 \\ lim_{x→o} − | x |= 0 \end{matrix}

Hence , by Sandwich Theorem , lim_{x→o} x sin \frac{1}{x} = 0

11. For every M > 0 , there exists δ > 0 such that

0 <| x − a |< δ ⇒ f(x) > M.