Limits, Continuity & Differentiability
Continuity And Differentiability
nta_abhyas_2025
Grade None
Question:
The function $f(x) = \frac{1 - \sin x + x}{1 - \sin x \cos x}$ is not defined at $x = \pi$. The value of $f(\pi)$, so that $f(x)$ is continuous at $x = \pi$, is
Step-by-Step Solution
Key Concept: Continuity requires left limit equals right limit equals function value at that point; use L'Hôpital's rule for indeterminate forms.
For continuity at $x = 0$, we need $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$. Computing $\lim_{x \to 0^-} f(x)$ using algebraic manipulation and L'Hôpital's rule: $\lim_{x \to 0} \frac{\sin(1-\cos x)}{1-\cos x} \cdot \lim_{x \to 0} \frac{x-\sin x}{x^2}$. The first limit equals $(1)^2 = 1$ and the second equals $\frac{1}{2}$, giving the left limit as $\frac{1}{2}$. Therefore $k = \frac{4}{5} \times 10k = \frac{4}{5} \times 10 \times \frac{1}{8} = 1.25$.
Correct Answer: 1