Question

Prove that the following function is continuous at {x=n}, where 'n' is a (+)ve integer :--

Answer 1

\[f(x)=x^n\]

Answer 2

well its derivatives are continuous

Answer 3

i dont know if that helps

Answer 4

What does it mean?

Answer 5

I mean derivatives of\[f(x)=x^n\]

Answer 6

\[f'(x)=nx^{n-1}\]

Answer 7

k!then, but how can we say that it is continuous?

Answer 8

If it's not defined at n then it's it's not continuous.

Answer 9

But it is continuous 100%

Answer 10

Did you prove it already?

Answer 11

no I didn't proved it ,but this is the question's demand!

Answer 12

Do you get an exact value fo any positive n?

Answer 13

Regardless of what positive you input you will always get an output.

Answer 14

no I got only \[f(n)=n^n\] where 'n' is a (+) integer

Answer 15

continuos

Answer 16

& question's demand is to prove it by limit method

Answer 17

we have to do just like\[\lim_{x \rightarrow n^-} f(x)=\lim_{x \rightarrow n^+} f(x)=f(n)\]
But I don't know how!

Answer 18

continuous because
\[\lim_{x\to a}x=a\] by sheer obviousness and therefore
\[\lim_{x\to a}x^n=a^n\] because the limit of a product is the product of the limits

Answer 19

Then can we say that \[a^n\] where a is any real no., is continuous.If yes then why? please tell.

Answer 20

as i understand it you have the function \(f(x)=x^n\) for some positive integer \(n\) and you want to show it is continuous (for any number \(a\))
this means you want to show that
\[\lim_{x\to a}f(x)=f(a)\] in this case it is
\[\lim_{x\to a}x^n=a^n\] that is what you need to show

Answer 21

k! now, can we show that f(x) is continuous at x=n from the above statement?

Answer 22

and i wrote above what ought to convince anyone that it is true, unless you need to use a "epsilon, delta" proof

Answer 23

well if it is true for all \(a\in \mathbb{R}\) is is certainly true for any \(n\)

Answer 24

k! thanx a lot.I got it at last.