Your problem is in the denominator [you need to get rid of ln(1+sin(x))], not in the numerator. Therefore, I don't think that will help you.Can You apply this to 4x 5x...nx?
Well the denominator îs x^2 becauseYour problem is in the denominator [you need to get rid of ln(1+sin(x))], not in the numerator. Therefore, I don't think that will help you.
hint: ln(1+x)→x when x→0.
Correct.Well the denominator îs x^2 because
x→0limsin(x)∗ln(1+sin(x))=
xsin(x)∗x∗sin(x)ln(1+sin(x)∗xsin(x)∗x=x2
You didn't answered my question @mario99
If you mean this by your question. I don't see how it helps. If you think that it helps, try it.Can I conjugate the square Root and cube Root using the formula a3−b3=(a−b)(a2+ab+b2)?
Conjugation?Correct.
Now rewrite your limit with this simplification:
x→0limsin(x)ln(1+sin(x))cos(x)−3cos(x)=x→0limx2cos(x)−3cos(x)
If the calculator has used conjugation, there must be a way to solve the problem by them, but I cannot see it. You are not allowed to use infinite series?Conjugation?
I used a limit calculator because they didn't teached us at highschool conjugation with hingher orders.
I'm clueless and we can't use differentiation because they didn't teached us.
Do You use both
a3−b3=(a−b)(a2+ab+b2) and
a3+b3=(a+b)(a2−ab+b2)?
The calculator used both of them.
I don't see how that will work, but I have an idea.So în my case a=cos(x) and b=3cos(x)?