Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade None
Question:
If $f(x) = \frac{e^{x^2}(1+4x)^{1/2}}{\ln(1-x^2)}$ for $x \neq 0$, then $f$ has:
An irremovable discontinuity at $x = 0$
A removable discontinuity at $x = 0$ and $f(0) = -4$
A removable discontinuity at $x = 0$ and $f(0) = -1/4$
A removable discontinuity at $x = 0$ and $f(0) = 4$
Step-by-Step Solution
Key Concept: Use Taylor series expansion of exponential and logarithmic functions to resolve $0/0$ indeterminate forms.
Evaluate $\lim_{x \to 0} \frac{e^{2x} - (1+4x)^{1/2}}{\ln(1-x^2)}$. Using Taylor series: numerator $= e^{2x} - (1+4x)^{1/2} \approx (1+2x+2x^2) - (1+2x-2x^2) = 4x^2 + O(x^3)$ and denominator $= \ln(1-x^2) \approx -x^2$. Thus the limit is $\frac{4x^2}{-x^2} = -4$.
Correct Answer: 2