Evaluating a Limit

[swift perch]

Are u allowed to use lhospitals rule for this problem?

[mellow stream]

@crisp grove Need help?

[wide basin]

the over x makes it undef

It's not some sin x over x thing but it's sin(3pi)/+-0

[crisp grove]

No it's forbidden to use this rule although it's the key to solve the problem

[swift perch]

Wdym by it’s forbidden to use but it’s key?

I don’t understand

[crisp grove]

My teacher will not be happy if use l'hopital

[icy nacelle]

you can technically say
$\lim_{x\to0}\frac{\sin{\left(\pi\sqrt{x^2+9}\right)}}{\pi\sqrt{x^2+9}}\frac{\pi\sqrt{x^2+9}}{x}$

[wind rivet]

Suika

[icy nacelle]

then the left part is 1

as it's a standard limit

then,

$\pi\sqrt{\lim_{x\to0}\frac{x^2+9}{x^2}}$

[wind rivet]

Suika

[icy nacelle]

then that is just $\pi$

[wind rivet]

Suika

[swift perch]

How is the 2nd part pi

I’m pretty sure that limit is DNE

Cuz as x approaches 0 from the left it approaches negative infinity

And as x approaches x from the right, it approaches positive infinity

Cuz it’s a vertical asymptote

[misty moth]

nor is the first part 1

The left part should be equal to 0, no?

As it goes to $\frac{sin(3\pi)}{3\pi} = 0$

[wind rivet]

rojan

[swift perch]

This part is 1

[misty moth]

How?

[swift perch]

But this part is DNE

[swift perch]

Cuz the limit of sin(x)/x as x goes to 0 is 1

[misty moth]

yes, but $\pi\sqrt{x^2 + 9}$ does not go to 0, x does.

[wind rivet]

rojan

[swift perch]

Oh wait

Mb

Ur right

But doing it this way still doesn’t work

Since the 2nd part is DNE

[misty moth]

what do that stand for again?

[swift perch]

This part

Is DNE

Does not exist

[misty moth]

It does exist, as x goes to 0 and is not equal to zero

(definition of limits)

[swift perch]

No the limit does not exist

[misty moth]

and as it exists and its multiplied by zero (the first part), the result is 0

[swift perch]

As it goes to 0 from the left, it goes to -inf, as it goes to 0 from the right, it’s positive inf

And 0*DNE is an indeterminate

[misty moth]

Yeah, makes sense

[misty moth]

okay, in the end you can solve it with a lot of manipulation involving $\sin{x+k\pi} = (-1)^k \sin{x}$ and $\lim_{x\to0} \frac{\sin(x)}{x} = 1$

[wind rivet]

rojan

[swift perch]

u can solve it with lhospitals rule

i feel like its cruel and unusual punishment to make someone do this limit without lhospitals rule 😭

[wide basin]

is any form of calculator allowed