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Epsilon delta proof with ln and sin(x)

0

1

$begingroup$

I tried it without using the fact |sin x| <= |x| for all x, but I know using this, would make it way easier. In fact, I'm not even sure if my proof is correct.

Proof:

Let ε > 0 be given. We want to find δ > 0 s.t. if $π/2$ < x < δ + $π/2$, then |ln(1+sin(x - $π/2$))| < ε.

|ln(1+sin(x - $π/2$))| < ε.

1 + sin(x - $π/2$)) < e^ε

arcsin(e^ε - 1) > x - $π/2$

δ = arcsin(e^ε - 1).

And since e^ε - 1 < 1, ε < ln 2. So, if given ε > ln 2, picking a δ for ε < ln 2.

So far, this is what I have. I'm pretty sure there should be a better way to do this. Thanks in advance for your help!

epsilon-delta

share|cite|improve this question

asked Mar 31 at 14:46

HellowhatsupHellowhatsup

305

$endgroup$

add a comment | 

0

1

$begingroup$

I tried it without using the fact |sin x| <= |x| for all x, but I know using this, would make it way easier. In fact, I'm not even sure if my proof is correct.

Proof:

Let ε > 0 be given. We want to find δ > 0 s.t. if $π/2$ < x < δ + $π/2$, then |ln(1+sin(x - $π/2$))| < ε.

|ln(1+sin(x - $π/2$))| < ε.

1 + sin(x - $π/2$)) < e^ε

arcsin(e^ε - 1) > x - $π/2$

δ = arcsin(e^ε - 1).

And since e^ε - 1 < 1, ε < ln 2. So, if given ε > ln 2, picking a δ for ε < ln 2.

So far, this is what I have. I'm pretty sure there should be a better way to do this. Thanks in advance for your help!

epsilon-delta

share|cite|improve this question

asked Mar 31 at 14:46

HellowhatsupHellowhatsup

305

$endgroup$

add a comment | 

$begingroup$

I tried it without using the fact |sin x| <= |x| for all x, but I know using this, would make it way easier. In fact, I'm not even sure if my proof is correct.

Proof:

Let ε > 0 be given. We want to find δ > 0 s.t. if $π/2$ < x < δ + $π/2$, then |ln(1+sin(x - $π/2$))| < ε.

|ln(1+sin(x - $π/2$))| < ε.

1 + sin(x - $π/2$)) < e^ε

arcsin(e^ε - 1) > x - $π/2$

δ = arcsin(e^ε - 1).

And since e^ε - 1 < 1, ε < ln 2. So, if given ε > ln 2, picking a δ for ε < ln 2.

So far, this is what I have. I'm pretty sure there should be a better way to do this. Thanks in advance for your help!

epsilon-delta

share|cite|improve this question

asked Mar 31 at 14:46

HellowhatsupHellowhatsup

305

$endgroup$

share|cite|improve this question

asked Mar 31 at 14:46

HellowhatsupHellowhatsup

305

share|cite|improve this question

asked Mar 31 at 14:46

HellowhatsupHellowhatsup

305

asked Mar 31 at 14:46

HellowhatsupHellowhatsup

305

asked Mar 31 at 14:46

HellowhatsupHellowhatsup

305