# Taylors Series for Limits

For the equation: $$\lim_{ x\to 1} \frac{1−x+\ln x}{1+\cos(\pi x)}$$

How can you evaluate this limit using a Taylor Series for both the numerator and deminator?

Would I need to create a taylor series for $ln x$ on the numerator and then another series for $\cos(\pi x)$? How does the rest of the equation become incorporated into this?

Thanks

• Make the change of variable $t = 1-x$, then $t \to 0$ as $x \to 1$. Then you can have a taylor series of $\log (1-t)$. – r9m May 16 '15 at 5:05

As said in answers and comments, setting $x=1-t$ gives $$\dfrac{1-x+\ln(x)}{1+\cos(\pi x)}=\dfrac{t+\ln(1-t)}{1+\cos(\pi-\pi t)}=\dfrac{t+\ln(1-t)}{1-\cos(\pi t)}$$ Now, using the usual expansions $$\log(1-t)=-t-\frac{t^2}{2}-\frac{t^3}{3}-\frac{t^4}{4}+O\left(t^5\right)$$ $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^5\right)$$ replace $y$ by $\pi t$ so $$\cos(\pi t)=1-\frac{\pi ^2 t^2}{2}+\frac{\pi ^4 t^4}{24}+O\left(t^5\right)$$ Then the numerator is $$-\frac{t^2}{2}-\frac{t^3}{3}-\frac{t^4}{4}+O\left(t^5\right)$$ and the denominator is $$\frac{\pi ^2 t^2}{2}-\frac{\pi ^4 t^4}{24}+O\left(t^5\right)$$ Now, perform the long division and the expression reduces to $$-\frac{1}{\pi ^2}-\frac{2 t}{3 \pi ^2}+O\left(t^2\right)$$ which not only shows the limit but also how it is approached.
• What I gave is the expansion of the expression around $t=0$. Plug $t=0$ to get the limit. Using more terms, Taylor would give around $x=1$ $$\frac{1-x+\log (x)}{1+\cos (\pi x)}=-\frac{1}{\pi ^2}+\frac{2 (x-1)}{3 \pi ^2}-\frac{\left(6+\pi ^2\right) (x-1)^2}{12 \pi ^2}+\cdots$$ – Claude Leibovici May 17 '15 at 3:23
Set $x=1-t$. We then have \begin{align} \lim_{x \to 1} \dfrac{1-x+\ln(x)}{1+\cos(\pi x)} & = \lim_{t \to 0} \dfrac{t+\ln(1-t)}{1+\cos(\pi-\pi t)} = \lim_{t \to 0} \dfrac{t+\ln(1-t)}{1-\cos(\pi t)} = \lim_{t \to 0}\dfrac{t-t-t^2/2+\mathcal{O}(t^3)}{2\sin^2(\pi t/2)}\\ & = \dfrac{-1/2}{2\cdot \pi^2/4} = -\dfrac1{\pi^2} \end{align}
Recall the usual limits $$\lim_{y\rightarrow 0}\frac{1-\cos y}{y^{2}}=\frac{1}{2},\ \ \ \ \ \ \lim_{y\rightarrow 0}\frac{\log (1+y)-y}{y^{2}}=-\frac{1}{2}.$$ Let us put $y=x-1,$ and recall that $\cos (\pi +\theta )=-\cos \theta ,$ then $$\lim_{x\rightarrow 1}\frac{1-x+\ln x}{1+\cos (\pi x)}=\lim_{y\rightarrow 0}% \frac{-y+\ln (1+y)}{1+\cos (\pi +\pi y)}=\lim_{y\rightarrow 0}\frac{\left( \frac{\ln (1+y)-y}{y^{2}}\right) }{\pi ^{2}\left( \frac{1-\cos (\pi y)}{(\pi y)^{2}}\right) }=\frac{-\left( \frac{1}{2}\right) }{\pi ^{2}\left( \frac{1}{2% }\right) }=-\frac{1}{\pi ^{2}}.$$ The usual limits used can be computed by L'Hospital's rule twice.