limits with trigonometry

[opal copper]

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[drowsy sky]

For x -> 0 we have:
sin(x) ~ x
1 - cos(x) ~ x^2/2
tan(x) ~ x

[small cedar]

??

[drowsy sky]

f(x) ~ g(x), x -> x0 means that the limit of f(x)/g(x) as x -> x0 is 1.
If you prefer power series, you can use it, too. For x -> 0:
sin(x) = x + o(x)
cos(x) = 1 - x^2/2 + o(x^2)
tan(x) = x + o(x)

[small cedar]

what does ~ stand for?

[drowsy sky]

Well, I just wrote what it means.

[small cedar]

oh

i'm kinda lost, i'm not sure what u were doing up there

[drowsy sky]

Let me show on one of the problems, then.
(a).
2sin(5x)/tan(2x) ~ 2(5x)/(2x) -> 5, x -> 0

[small cedar]

where did sin and tan go?

[drowsy sky]

As I said above, for x -> 0 we have sin(x) ~ x, tan(x) ~ x. So, sin(5x) ~ 5x and tan(2x) ~ 2x for x -> 0.

[small cedar]

ohh

[small cedar]

so the answer is just 5?

[drowsy sky]

You need to use some terms from the series of trigonometric and some other functions to solve limits with them. So, I recommend reviewing the series for the following:
(1 + x)^n
ln(1 + x)
e^x
sin(x)
cos(x)
tan(x)

[drowsy sky]

Yes.

[small cedar]

wait but u said that sin(x) = x, so 2 sin (5x) would be 2x(5x)?

[drowsy sky]

You can't say sin(x) = x. For x -> 0 we have sin(x) ~ x, so 2sin(5x) ~ 2(5x) = 10x, x - > 0.

[small cedar]

i don't get it..

[drowsy sky]

Well, sorry, I don't know how to explain it in the other way. Review some theory about Taylor series, then.

And about Landau symbols so you can calculate limits more quickly.

[small cedar]

oh ok then, thanks