Question

Please help me in this limit question...

Answer 1

question you must ask,
help we will

Answer 2

Let \[ f(x)=\begin{cases}1 &if \ x=\frac{1}{n} \ where \ n\epsilon \mathbb{Z}^{+}\\
0 & \mbox{otherwise}\end{cases}\]
(i) Show that \[c\neq 0\] then \[\lim_{x \to c}f(x)=0\]
(ii) Show that \[\lim_{x \to 0}f(x)\] does not exist.

Answer 3

@thefaceless

Answer 4

@eliassaab

Answer 5

\[ Thanks \ Ganeshie \]

Answer 6

\[\begin{equation}
\lim_{x \to c}f(x)=0 \ \ when \ c=1, \frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots,\frac{1}{n}, \ldots
\end{equation}
\\Above \ can\ be\ proved\ easilly.\\When\ c\ is\ any\ value\ other\ than\ 0,\\it\ can\ be\ proved\ that,\]

Answer 7

\[\begin{equation}
\lim_{x\to c}f(x)=0
\end{equation}\\\mbox{The problem arises when c=0}\]
Why this happens because, c=0 means,

Answer 8

\[\begin{equation}
c=\lim_{n \to \infty }\frac{1}{n} \mbox{where n is a positive integer.}
\end{equation}
\]
Or,

Answer 9

\[
\begin{equation}
c=\lim_{m \to \infty}\frac{1}{m} where\ m\notin \mathbb{Z}^{+}
\end{equation}
\]

Answer 10

\[
\mbox{This } \infty \mbox{ makes the problem here. We can't say whether $\infty $ is a integer or not. }
\]

Answer 11

not.

Answer 12

Thus, even,
\[
\mbox{$\lim_{x \to 0^{+}}f(x)$ does not exist.}

\]

Answer 13

(because it can be both 1 and 0)

Answer 14

As the right limit does not exist we can say, \[
\lim_{x \to 0}{f(x)}\mbox{ does not exist}
\]

Answer 15

I don't know whether my above arguments are correct.
Please help me to prove this.

Answer 16

sorry man there is something wrong with my equation icon but in question (ii) you should use the property of continuity which says the limit of the function f(x) when x approaches zero is equal to f(0).

Answer 17

@Percy* I haven't heard of it. I'll check it. Thank you! BTW I used latex :-)

Answer 18

can you present them in one entry @amilapsan ?

Answer 19

@nincompoop what do you mean by "them"?

Answer 20

Suppose that \( c\ne 0\), choose N so that n > N implies \( \frac 1 n \)< |c| . This implies that finite numbers of of \( \frac 1 n \) will be near c, so f(x)= 1 except for finitely many x, hence the limit of f(x) is 1 when x approaches c

Answer 21

Take \( x_n=\frac 1 n\to 0\) and \( y_n=\frac{\sqrt 2}n\to 0\) but
\( f(x_n)=1 \) and \(f(y_n)=0\) so the limit of f(x) does not exist when x goes to 0

Answer 22

@ganeshie8