Limits

[atomic owl]

How can I solve the limit as x approaches infinity for (sin2x)/x?

[languid terrace]

by using squeeze theorem or sandwich theorem

sin2x belongs to [-1, 1]

multiply the inequation with 1/x

youll get -1/x </= sin(2x) / x </= 1/x

by squeeze theorem youll need to evaluate the limits of -1/x and 1/x as x approaches infinity

which is zero

[atomic owl]

Do I always just test 1/x and -1/x?

Whenever I’m using the Squeeze Theorem?

[languid terrace]

for this function it is 1/x and -1/x

for others it might be different

[atomic owl]

How do you know? Is that something I have to have memorized?

[languid terrace]

nope not at all

just find the range of the given function

[atomic owl]

How do I do that?

[languid terrace]

like we just did

sinx always lies in [-1,1]

[atomic owl]

Oh, okay

[languid terrace]

any value of x you put in, it always gives a result in between those values

so you set up an inequation

like we just did

and try to make the function the question is asking

which we can just do by multiplying both the sides of the inequation with 1/x

[atomic owl]

I see

[languid terrace]

the rule of thumb for squeeze theorem is you have to know a function which is always less than the function whose limit you are trying to evaluate, and a function which is always greater than the given function

for the same x of course

[atomic owl]

Alr

Tysm

[languid terrace]

np

here

that should sum it up

here g(x) is the given function

[blazing folio]

.solved