GATE CS 1993 Q02.1 — Calculus: Limits, Continuity, Differentiability

To evaluate the limit, we can use the Taylor series expansions of $e^x$ and $\cos x$ around $x = 0$:

• $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \mathcal{O}(x^4) \implies e^x - 1 = x + \frac{x^2}{2} + \frac{x^3}{6} + \mathcal{O}(x^4)$
• $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \mathcal{O}(x^6) \implies \cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} - \mathcal{O}(x^6)$

Now, substitute these expansions into the numerator and the denominator:

Numerator:
$$x(e^x - 1) + 2(\cos x - 1) = x\left(x + \frac{x^2}{2} + \mathcal{O}(x^3)\right) + 2\left(-\frac{x^2}{2} + \mathcal{O}(x^4)\right)$$

$$= \left(x^2 + \frac{x^3}{2} + \mathcal{O}(x^4)\right) - x^2 + \mathcal{O}(x^4) = \frac{x^3}{2} + \mathcal{O}(x^4)$$

Denominator:
$$x(1 - \cos x) = x\left(\frac{x^2}{2} - \mathcal{O}(x^4)\right) = \frac{x^3}{2} - \mathcal{O}(x^5)$$

Substituting these back into the limit expression:

$$\lim_{x \to 0} \frac{\frac{x^3}{2} + \mathcal{O}(x^4)}{\frac{x^3}{2} - \mathcal{O}(x^5)} = \lim_{x \to 0} \frac{\frac{1}{2} + \mathcal{O}(x)}{\frac{1}{2} - \mathcal{O}(x^2)} = \frac{1/2}{1/2} = 1$$