Question:

Step 1: Verify the form at $x = 0$ The function is given by $f(x) = \frac{x - e^x + \cos 2x}{x^2}$. At $x = 0$, the numerator is $0 - e^0 + \cos(0) = 0 - 1 + 1 = 0$. The denominator is $0^2 = 0$. Thus, the expression is of the indeterminate form $\frac{0}{0}$. To find $f(0)$ for continuity, we must evaluate this limit. Step 2: Apply Taylor series expansions around $x = 0$ The Taylor series expansions for $e^x$ and $\cos(2x)$ around $x=0$ are: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^4)$$ $$\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - O(u^6)$$ Substituting $u=2x$ into the expansion for $\cos(u)$: $$\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - O(x^6) = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} - O(x^6) = 1 - 2x^2 + \frac{2x^4}{3} - O(x^6)$$ Step 3: Expand the numerator Substitute the Taylor series expansions into the numerator of $f(x)$: $$x - e^x + \cos(2x) = x - \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right) + \left(1 - 2x^2 + O(x^4)\right)$$ $$= x - 1 - x - \frac{x^2}{2} - \frac{x^3}{6} + 1 - 2x^2 + O(x^4)$$ Combine like terms: $$= (-1 + 1) + (x - x) + \left(-\frac{x^2}{2} - 2x^2\right) - \frac{x^3}{6} + O(x^4)$$ $$= 0 + 0 + \left(-\frac{1}{2} - \frac{4}{2}\right)x^2 - \frac{x^3}{6} + O(x^4)$$ $$= -\frac{5}{2}x^2 - \frac{x^3}{6} + O(x^4)$$ Step 4: Calculate the limit For $f(x)$ to be continuous at $x=0$, $f(0)$ must be equal to $\lim_{x \to 0} f(x)$. $$f(0) = \lim_{x \to 0} \frac{-\frac{5}{2}x^2 - \frac{x^3}{6} + O(x^4)}{x^2}$$ Divide each term in the numerator by $x^2$: $$f(0) = \lim_{x \to 0} \left(-\frac{5}{2} - \frac{x}{6} + O(x^2)\right)$$ As $x \to 0$, the terms containing $x$ or higher powers of $x$ approach zero: $$f(0) = -\frac{5}{2}$$