The first message. Also a test.

← 705 messages moved from In pursuit of moksha

Yo

Apparently there is a feature that allows you to move messages from one room to another, if you're a room owner.

So, we've still got the previous transcripts with us in this new room

@ViditD How's life going, bro? :)

I'm awaiting your reply to my "Letters" eagerly :)

Ohh, it's tomorrow? Sahi, all the best bro. Do well (Y)

What's the center for the exam?

I see

How's the prep? Feeling good about it?

:)

@ViditD Didn't follow, what do you mean bro?

@ViditD Ahh, okay

Not really

I don't think SE chat has received any improvements over many years

But, I vaguely recall some improvement was made to the mobile UI? Not sure, though

@toric_actions Yo bro

@toric_actions How's the weather? :D

Uggh...what happened? :(

@ViditD +1. Snowing?

(Oh wait, there's a starring system to +1, totally forgot about that.)

Yup. Still keeps happening to me once every year, like a ritual :/

@toric_actions Yeah, but it can double up as a +1-ing system too

@ViditD True

@toric_actions Do you have to handle your own medical expenses?

@toric_actions Ohh I see

@toric_actions I don't think so, but maybe "Let us know how the exam tomorrow goes" is more accurate.

@ViditD May no such disorders come your way!

@toric_actions What's he up to these days? :)

Oh, interesting

Is IISER Pune the only IISER with separate TA positions?

@ViditD The only problematic point is $x = 0$, so let us check there. If $f(x)$ is the given function, then for $h > 0$ we have $(f(h) - f(0)) / h = \frac{\sin(h)}{h(h + \cos(h))} = \frac{\sin(h)/h}{h + \cos(h)}$. So, in the limit $h \to 0^+$, we have that $(f(h) - f(0))/h \to 1$.

@ViditD Similarly, let $h > 0$ and consider $(f(-h) - f(0))/(-h)$. This equals $\frac{\sin(-h)}{-h(h + \cos(-h))} = \frac{\sin(h)/h}{h + \cos(h)}$. Again, in the limit $h \to 0^+$, we have that $(f(-h) - f(0))/(-h) \to 1$.

Since the left and right hand limits exist and are equal, the given function $f$ is differentiable at $x = 0$. Hence, it is differentiable everywhere.

@ShubhamNamdeo Yo bro