Question

Value of limx → 0⁡(1+Sin(x))Cosec(x)​

Answer 1

$\Huge\bf\maltese{\underline{\green{Answer°᭄}}}\maltese$

$\implies$$\large\bf{\underline{\red{VERIFIED✔}}}$

A function is continuous if the preimage of every open set is an open set. This is equivalent to requiring that the preimage of every closed set be closed (the things that get mapped to the complement of your open set, are exactly the complement of the things that got mapped to your open set).

With this in mind, suppose ∃x:f(x)≠f(0)∃x:f(x)≠f(0). Then f−1({f(x)})⊃{x,x/2,x/4,x/8,…}

f−1({f(x)})⊃{x,x/2,x/4,x/8,…} but it does not contain 0. So the closed set {f(x)}{f(x)} does not have a closed preimage, and the function is not continuous.

$\boxed{I \:Hope\: it's \:Helpful}$

${\sf{\bf{\blue{@ℐᴛz ᴅɪɴᴜ࿐}}}}$

Answer 2

Answer:

$\Huge\bf\maltese{\underline{\green{Answer°᭄}}} \bf \maltese$

$\implies\large\bf{\underline{\red{VERIFIED✔}}}$

A function is continuous if the preimage of every open set is an open set. This is equivalent to requiring that the preimage of every closed set be closed (the things that get mapped to the complement of your open set, are exactly the complement of the things that got mapped to your open set).

With this in mind, suppose ∃x:f(x)≠f(0)∃x:f(x)≠f(0). Then f−1({f(x)})⊃{x,x/2,x/4,x/8,…}

f−1({f(x)})⊃{x,x/2,x/4,x/8,…} but it does not contain 0. So the closed set {f(x)}{f(x)} does not have a closed preimage, and the function is not continuous.

$\boxed{I \:Hope\: it's \:Helpful}$

$\blue{@Pratyushara987࿐}$