someone can help me solving this limit ? (without L'Hospital rule)

[warm flame]

here is the limit

[tulip totem]

Expand all three using power series.

[warm flame]

not allowed

we can only use this rules sinx/x and tanx/x

[tulip totem]

I see.

Divide the top and bottom by x

And then multiply the top and bottom by 1+cos(x)

[dapper pollen]

Why not lopital rule?

[tulip totem]

And then use 1 - cos^2(x) = sin^2(x)

[warm flame]

it's not allowed in my grade i asked the teacher about that so

[dapper pollen]

Ok give me a sec doing things in paper

[warm flame]

thank u

@tulip totemi'll try doing what u said

[hexed gate]

@warm flame has given 1 rep to @dapper pollen

[tulip totem]

you get:

(1+cos(x))(tan(5x) + 2(sinx/x))
————
(sinx)(sinx/x)

[dapper pollen]

Wth

[warm flame]

why not x^2 ?

so like : (1 - cosx)/x^2 = 1/2

[tulip totem]

Yes.

[warm flame]

it's good ?

[tulip totem]

No, because for some reason, the method I suggested fails.

[dapper pollen]

i figured out how to do it

[tulip totem]

Wait. No. The limit is divergent.

[warm flame]

yeees tell me

[dapper pollen]

Well, kinda easy

[warm flame]

okay

[compact siren]

No need.
First, use the double-angle formula for cosine and divide everything by x:
(x tan(5x) + 2sin(x))/(1 - cos(x)) = (tan(5x) + 2sin(x)/x)/(2sin(x/2)^2/x)
Now it's easy to verify that the limit doesn't exist.

[warm flame]

okay wait

no ideas ?