Question:

Step 1: Expand each term using Taylor series around $x=0$. The Taylor series for $e^x$ is: $$e^x = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$ The Taylor series for $\sin 2x$ is: $$\sin 2x = (2x) - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \dots = 2x - \frac{4x^3}{3} + \dots$$ The Taylor series for $\sqrt{1-x}$ is: $$\sqrt{1-x} = (1-x)^{1/2} = 1 + \frac{1}{2}(-x) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(-x)^2 + \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}(-x)^3 + \dots$$ $$\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \dots$$ Step 2: Substitute these expansions into the numerator and collect terms by powers of $x$. $$ae^x+b\sin 2x+c\sqrt{1-x} = a\left(1+x+\frac{x^2}{2}+\dots\right)+b\left(2x-\frac{4x^3}{3}+\dots\right)+c\left(1-\frac{x}{2}-\frac{x^2}{8}-\dots\right)$$ $$= (a+c) + \left(a+2b-\frac{c}{2}\right)x + \left(\frac{a}{2}-\frac{c}{8}\right)x^2 + \dots$$ Step 3: For the limit $\lim_{x \to 0} \dfrac{ae^x + b\sin 2x + c\sqrt{1-x}}{x^2}$ to exist finitely, the numerator must have $x^2$ as its lowest power term. This requires the constant term and the coefficient of $x$ in the numerator to be zero. Step 4: Set the constant term to zero. $$a+c=0$$ This implies $c=-a$. Step 5: Set the coefficient of $x$ to zero. $$a+2b-\frac{c}{2}=0$$ Substitute $c=-a$ from the previous step into this equation: $$a+2b-\frac{(-a)}{2}=0$$ $$a+2b+\frac{a}{2}=0$$ $$\frac{3a}{2}+2b=0$$ $$3a+4b=0$$ Step 6: The conditions for the limit to exist finitely are: $$a+c=0$$ $$3a+4b=0$$