Question

\[f(x)=\left\{\begin{matrix}
sinx:x\neq n\pi ,n\in \mathbb{Z}\\0,otherwise &
\end{matrix}\right.\
g(x)=\left\{\begin{matrix}
x^2+1, if x\neq 0,2\\x, if x=0\\5, if x=2
\end{matrix}\right.\] then \[\lim_{x \to 0}g[f(x)]\]=

Answer 1

Hmm ...
I think you can do this
\[ \lim_{x \rightarrow 0} f(g(x)) = \lim_{x \rightarrow 0} f\left ( \lim_{x \rightarrow 0}g(x) \right )\]

Answer 2

And how do we know if \(x^2\)+1 is equal to or not equal to n\(\pi\) ?

Answer 3

OS... tyestai ho.. F5 helps.

Answer 4

Hmm ...
g(x) = x^2 + 1 as x->0 which tends to 1

Answer 5

which is \( \neq \) nπ

Answer 6

still ... i'm not so sure ...

Answer 7

Isnt it g(f(x))?

Answer 8

Ah! lets begin again.

Answer 9

Oh ... my bad ;(

Answer 10

i guess it would be zero in both cases

Answer 11

for f(x)

Answer 12

My bad x NOT equal to npi. :/

Answer 13

So its,
\[\lim_{x \rightarrow 0} \sin^2 x + 1\]
Which is basically 1.

Answer 14

Here x tends to 0 implies x tends to 0\(\pi\) and is not equal to n\(\pi\) so, f(x)= sinx
and then g(sinx)= \(\sin^2 x\)+1.
=1 (putting x=0)

Is that correct?

Answer 15

Yesss. :)

Answer 16

thanks!