Question

discuss it s continuity

Answer 1

It's mostly a rational function, which is continuous everywhere on its domain, right?

Answer 2

Try to get the one-sided limit of the rational function, as x goes to 1

Answer 3

Maybe this would help....
\[\large 1-x^n = (1-x)(1+x+x^2+...+x^{n-1})\]

Answer 4

okay let me try

Answer 5

hmm
not getting

Answer 6

So that you can cancel out the denominator
\[\large \frac{1-x^n}{1-x}=\frac{\color{red}{(1-x)}(1+x+x^2+...+x^{n-1})}{\color{red}{1-x}}\]

Answer 7

yes i did it but what after that

Answer 8

Well, after the denominator is cancelled, you can get the limit as x goes to 1, much more straightforward-ly
\[\Large \lim_{x\rightarrow 1}(1+x+x^2+...+x^{n-1})\]

Answer 9

Having doubts? All polynomials are continuous.

Answer 10

now, i have to calculate LHL and RHL of it

Answer 11

Why bother? It's a polynomial, it's continuous everywhere... but if you insist
(Just replace the x's with 1's... have mercy on yourself LOL )

Answer 12

yes, of course

Answer 13

And you'd get?

Answer 14

but answer is discontinuous

Answer 15

Of course... and to show that, we have to show that the limit of the function, f(x) as x goes to 1 is not equal to f(1)

That alone suffices to show discontinuity.

Answer 16

okay
@terenzreignz thanks bro

Answer 17

Are you SURE you got it from here?

Answer 18

hope so

Answer 19

Well, should you have any doubts.... this site stands ready to allay them :)

Answer 20

LHL ;limx→1(1+x+x^2+...+x^n−1)
RHL ; n-1
LHL is not equal to RHL

Answer 21

Okay.... wrong approach...
LHL = RHL all right, because
\[\huge f(x) = \left\{\begin{array}{11}\frac{1-x^n}{1-x}\quad\quad x\ne 1\\ \\ \\ \\ \\ n-1 \quad\quad\quad \ x=1 \end{array} \right. \]

Answer 22

so, in actuality...
\[\huge f(x) = \left\{\begin{array}{11}\frac{1-x^n}{1-x}\quad\quad x< 1\\ \\ \\ \\ \\ n-1 \quad\quad\quad \ x=1 \\ \\ \\ \\ \frac{1-x^n}{1-x} \quad \quad x > 1 \end{array} \right.\]

Answer 23

So whether you approach 1 from the left or right, you approach it with the same function.
It'is discontinuous, NOT because the LHL and RHL are unequal (they are equal) but because
\[\huge \lim_{x\rightarrow 1 }f(x) \ne f(1)\]

Answer 24

okay
i got it

Answer 25

Good. That's all I wanna know :)