Question

{an}= {f(n)} 1) Could there be a sequence limitz400 f (a) exists, but such that limit noo An does not? 2) Could limit nGoo Un exists but not for limit;f (2) 3) Discuss your options , you can give an example to enhance your reasoning.

Answer 1

1) NO, there could be no such sequence.

Proof:

lim f(0) erists and lim f(1) = a 1+00 1+00 → Given e > 0,3 MeR such that a - f(1)<, Vx > M

$Now\: notice\: that\: for\: all\: natural\: numbers\: n\geqslant M,$

a-f(n)<E, VneN such that n >M

$Notice\:that\:the\:above\:statement\:is\:the\:convergence\:criterion\:for\left \{ a_{n} \right \}.$

$\Rightarrow The\:sequence\:\left \{a _{n} \right \}=\left \{ f(n) \right \} is \:convergent \:and\: converges\:to\:a.$

Hence, proved.

2) YES,such a function exists.

$Example:\\ f(x)=\begin{cases} 1& \text{ if } x\:\epsilon\:\mathbb{N} \\ 0& \text{ otherwise } \end{cases}$

$Here, \:f(n) = 1 \:where \:n\:\epsilon\: \mathbb{N}\\$

$\Rightarrow \left \{ a_{n} \right \}= 1,1,1..... which\:is\:a\:constant\:sequence.\\ Thus,\lim_{n\rightarrow \infty }\left \{ a_{n} \right \} exists\: and \:\lim_{n\rightarrow \infty }\left \{ a_{n} \right \}=1\\$

$Clearly \: \lim_{x\rightarrow \infty } f(x) does\:not\:exist$$It\:is\:because\:it\:attains\:the\:value\:1\:at\:infinitely \:many\:points\:as\: x\rightarrow \infty$ .

3) We have already discussed an example for Question 2.

We are now about to discuss an example to support our proof of Question 1.

$Example:\\ f(x)= \frac{1}{x} \:\:\:for\:x\geqslant 1 \\ \lim_{x\rightarrow \infty }f(x)\:exists \:and \lim_{x\rightarrow \infty }f(x)=0.\\$

$\left \{ a_{n} \right \}=\left \{ f(n) \right \} = \left \{ \frac{1}{1} ,\frac{1}{2},\frac{1}{3},\frac{1}{4}...\right \}$

$Here, \lim_{n\rightarrow \infty }a_{n}\:exists\:and\:is\:equal\:to\:0.$

The formal proof is already shown in Question 1.

Hence, the above example supports the proof.

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