Find the limit without using L'Hospitall Rule

Question

Evaluate:$$\displaystyle \lim_{x \rightarrow 0}\frac{\cos(1-\frac{\sin x}{x})+\cos(2-\frac{\sin(2x)}{x})+\cdots+\cos(k-\frac{\sin(kx)}{x})-k}{\cos(1-\cos(x))+\cos(2-2\cos(2x))+\cdots+\cos(k-k\cos(kx))-k}$$

I can find this limit using L' Hospital Rule, I do not know how to do it without that.This question is proposed to Romanian Math Magazine by Jalil Hajimir

Answer 1

Let's observe that $$\frac{1-\cos(j-j\cos jx)} {j^2(1-\cos jx) ^2}\to\frac{1}{2}$$ and hence $$\frac{1-\cos(j-\cos jx)} {x^4}=\frac{1-\cos(j-\cos jx)} {j^2(1-\cos jx) ^2}\cdot j^2\cdot\frac{(1-\cos jx)^2} {(jx)^4}\cdot j^4\to \frac{j^6}{8}$$ Thus if we change sign of both numerator and denominator of the expression under limit in question and divide both numerator and denominator by $x^4$ then the denominator of resulting expression tends to $\sum_{j=1}^{k}j^6/8$.

To deal with numerator let's observe that a typical term of numerator is $$\dfrac{1-\cos\left(j-\dfrac{\sin jx} {x} \right)} {x^4}=\dfrac{1-\cos\left(j-\dfrac{\sin jx} {x} \right)} {\left(j-\dfrac{\sin jx} {x} \right) ^2}\cdot\frac{(jx-\sin jx) ^2} {(jx) ^6}\cdot j^6\to\frac{j^6}{72}$$ and hence the numerator tends to $\sum_{j=1}^{k}j^6/72$. The desired limit is thus $1/9$. Note that we need here the result that $$\frac{x-\sin x} {x^3}\to\frac{1}{6}$$ which is easily handled via Taylor series.

Answer 2

Let proceed by Taylor’s series to obtain

$$\cos\left(k-\frac{\sin(kx)}{x}\right)=\cos\left(k-\frac1x\left(kx-\frac16k^3x^3+o(x^3)\right)\right)=$$

$$=\cos\left(\frac16k^3x^2+o(x^2)\right)=1-\frac1{72}k^6x^4+o(x^4)$$

and

$$\cos(k-k\cos(kx))=\cos\left(k-k\left(1-\frac12k^2x^2+o(x^2)\right)\right)=$$$$=\cos\left(\frac12k^3x^2+o(x^2)\right)=1-\frac18k^6x^4+o(x^4)$$

then

$$\frac{\sum_{i=1}^k \frac{i^6x^4}{72}+o(x^4)}{\sum _{i=1}^k \frac{i^6x^4}{8}+o(x^4)}= \frac{\sum_{i=1}^k \frac{i^6}{72}+o(1)}{\sum _{i=1}^k \frac{i^6}{8}+o(1)}\to \frac19 \frac{\sum_{i=1}^k i^6}{\sum _{i=1}^k i^6}=\frac19$$