Differential Calculus-1
Differential Calculus-1
Allen Star Batch
Grade 12
Question:
If $f(x) = \begin{cases} \frac{x \cdot \ln(\cos x)}{\ln(1+x^2)} & x \neq 0 \\ 0 & x = 0 \end{cases}$ then:
$f$ is continuous at $x = 0$
$f$ is continuous at $x = 0$ but not differentiable at $x = 0$
$f$ is differentiable at $x = 0$
$f$ is not continuous at $x = 0$
Step-by-Step Solution
Key Concept: Evaluate continuity at x=0 using L'Hôpital's rule on the limit of x·ln(cos x)/ln(1+x²), then determine differentiability by computing left and right derivatives using Taylor expansions: ln(cos x) ≈ -x²/2 - x⁴/12 and ln(1+x²) ≈ x² - x⁴/2.
For the right derivative: $f'(0^+) = \lim_{h \to 0^+} \frac{h \ln(\cosh h)}{h \cdot \frac{\ln(1+h^2)}{h^2}} = \lim_{h \to 0^+} \frac{h^2 \ln(\cosh h)}{\ln(1+h^2)} = \lim_{h \to 0^+} \frac{-\sinh}{(\cosh) \times 2h} = \frac{1}{2}$. For the left derivative: $f'(0^-) = \lim_{h \to 0^-} \frac{-h \ln(\cosh h)}{-h \cdot \frac{\ln(1+h^2)}{h^2}} = -\frac{1}{2}$. Since the right and left derivatives are not equal, the function is not differentiable at $x=0$.
Correct Answer: 1,3