Calc

[slow olive]

Do you know L'Hopital's Rule?

You can apply L'Hopital's Rule repetitively if applying it once still results in an indeterminate form.

[marsh trellis]

L'Hôpital's twice is necessary

[slow olive]

Lhopital twice solves the question.

[marsh trellis]

-½

[terse otter]

@slow olive @marsh trellis just multiply top and bottom by cosx+1 and factor out a sinx/x

@rain bear

[marsh trellis]

@terse otter good spot

-1/(1+cosx) × sinx/x

-½

[odd lagoon]

What does it mean if f is continuous on (-pi, pi)?

[odd lagoon]

It's not.

[marsh trellis]

if you're doing L'Hôpital's then you need to do L'Hôpital's twice from that point

or you can do what trig identity said

[odd lagoon]

That's what I meant. L'Hopital's twice isn't necessary because L'Hopital's once isn't necessary.

[marsh trellis]

yes

[odd lagoon]

But also I think you guys might be skipping ahead of where their confusion is.

[rain bear]

the function crosses on -pi, pi

[odd lagoon]

...what?

[rain bear]

is that not what continuous means?

[marsh trellis]

nah

continuous means it undergoes no jumps, interruptions or breaks on that interval

also (and an important note) is that it is defined on every point in that interval

which means that the limit for f(x) does exist

[marsh trellis]

I read through the question again, and f(x) is that function providing x≠0, but based on the information that you have been told (that it is continuous on that interval) you know that the value of the function at x=0 is exactly what that function approaches on either side of 0

[odd lagoon]

For a function f to be continuous on an interval (a, b), it must be continuous on every point within that interval. And the definition of continuity at a point is that a function f is continuous at a point x = a if and only if lim_(x -> a) f(x) = f(a).

[terse otter]

Yo someone help me. Do the natural numbers include 0 or not what’s the standard

[odd lagoon]

The standard is to not hijack someone else's post.

[sweet wharf]

Make a post with that question

[empty cedar]

The function is continuous on that interval

So the limit needs to be equal to the value

It already exists

So we don't need to worry about that

Using l'hopitale is not good practice

U can def just solve this using trig identities

$\frac{\cos(x) -1}{x\sin(x)}
= \frac{\cos^2(x) - 1}{x\sin(x)(\cos(x) + 1)}
=- \frac{\sin(x)}{x(\cos(x)+1)}$

[slim mica]

Arm

[empty cedar]

U can easily compute that limit