Finding the $\lim \limits_{x \to 0} {1 - \cos(x)\over \sin(x) \ln(1+x)}$ using Taylor's series.

Question

I am a bit stuck. This is what I have so far and I am not sure how to simplify it further:

$${{x^2\over 2} - o(x^4)\over (x - {x^3 \over 6} + o(x^5))(x - {x^2 \over 2} +o(x^3))} $$

How do I proceed further?

Answer 1

What you did is absolutely correct and i hope that i can start from there $$\lim_{x\to o}\frac{1-cosx}{sin(x)ln(1+x)}=\lim_{x\to 0}\frac{\frac{x^2}{2}-\frac{x^4}{24}}{(x-\frac{x^3}{6})(x-\frac{x^2}{2}+\frac{x^3}{3})}$$ now take $x^2$ common in both numerator and denominator(one $x$ from each braces in the denominator) and then check the coefficient of constant in both numerator and denominator which gives you the value of limit (since all other powers of x tend to zero as x tends to zero) and in this case the value of limit is $$\lim_{x\to o}\frac{1-cosx}{sin(x)ln(1+x)}=\frac{1}{2}$$ Hope that helps!

Answer 2

Do some asymptotic analysis, it greatly simplifies things:

• $1-\cos x\sim_0 -\dfrac{x^2}2$;
• $\sin x\sim_0 x$, $\;\ln(1+x\sim_0x$.

$$\text{Therefore }\qquad\qquad{1 - \cos(x)\over \sin(x) \ln(1+x)}\sim_0\frac{-\cfrac{x^2}2}{x\cdot x}=-\frac12.$$

Answer 3

Don't go that far !

With some training, you notice that you essentially have

$$\frac{1-1+\dfrac{x^2}2}{x\cdot x}.$$

With rigor,

$$\frac{\dfrac{x^2}2+o(x^2)}{(x+o(x))(x+o(x))}=\frac{\dfrac{x^2}2+o(x^2)}{x^2+o(x^2)}\to\frac12.$$