help determining $a,b$ such that limit will exist and calculating it

Question

So i have a limit: $$\lim_{x\rightarrow 0} \frac{e^{ax}+b\ln(2-x)-1}{x-\sin{x}}$$

And i need to determine values for $a,b$ so that the limit will exist, and calculate the limit then. There is also a $hint$, that i need to use Taylor series for this, so it might be helpful, but i don't know if i should look for the whole function or just a part of fraction. Any help would really be appreciated. Thank you in advance.

Answer 1

$$\lim_{x\rightarrow 0} {x-\sin{x}} = 0$$

In order to make the limit exist, the numerator of fraction $\frac{e^{ax}+b\ln(2-x)-1}{x-\sin{x}}$ must be zero so that the limit will become undefined form.

$$\lim_{x\rightarrow 0} e^{ax}+b\ln(2-x)-1 = 1+ b \ln(2) -1=0$$ Then use L'hosptial rule, $$\lim_{x\rightarrow 0} \frac{e^{ax}-1}{x-\sin{x}}=\lim_{x\rightarrow 0} \frac{ae^{ax}}{1-\cos{x}}$$

To make the limit exist, in the same way, $\lim_{x \to 0} ae^{ax} = a = 0$

After L'hosptial twice, the answer should be $\lim_{x\rightarrow 0} \frac{e^{ax}+b\ln(2-x)-1}{x-\sin{x}} = 0$ and $a=b= 0$

Answer 2

As suggested use Series Expansion,

$$e^{ax}=1+\dfrac{ax}{1!}+\dfrac{(ax)^2}{2!}++\dfrac{(ax)^3}{3!}+\cdots$$

$$\ln(2-x)=\ln2+\ln\left(1-\dfrac x2\right)=\ln2-\sum_{r=1}^\infty\dfrac{(x/2)^r}r$$

$$\sin x=x-\dfrac{x^3}{3!}+\cdots$$

Answer 3

It is sure that Taylor series help a lot. Let us try $$e^{ax}=1+a x+\frac{a^2 x^2}{2}+\frac{a^3 x^3}{6}+O\left(x^4\right)$$ $$\log(x-2)=\log (2)-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{24}+O\left(x^4\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Replacing $$\frac{e^{ax}+b\ln(2-x)-1}{x-\sin{x}}=\frac{b \log (2)+x \left(a-\frac{b}{2}\right)+x^2 \left(\frac{a^2}{2}-\frac{b}{8}\right)+x^3 \left(\frac{a^3}{6}-\frac{b}{24}\right)+O\left(x^4\right)}{\frac{x^3}{6}+O\left(x^4\right)}$$ In order the limit exists, the constant term and the coefficients of the first and second powers of $x$ in numerator must be $0$. Then $???$