Question

The value of \( \lim_{x \to 0} \frac{(1 + x)^{\frac{1}{x}} - e + \frac{1}{2}e^x}{x^2} \) is:

• A. \(\frac{11}{24e} \)
• B. \(-\frac{11}{24e} \)
• C. \(\frac{e}{24} \)
• D. None of these

Hint:

For limits involving complex functions, expansions such as Taylor or Maclaurin series are extremely helpful in simplifying expressions by removing higher order terms which do not affect the limit.

Answer 1

The Correct Option is A

Let \( y = (1 + x)^{\frac{1}{x}} \). Then, \[ \log y = \frac{1}{x} \log(1 + x) = \frac{1}{x} \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \right) \] \[ = 1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \ldots \] So, \[ y = e^{\log y} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \ldots} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - \ldots} \] \[ = e \left( 1 + \left( -\frac{x}{2} + \frac{x^2}{3} \right) + \frac{1}{2!} \left( -\frac{x}{2} + \frac{x^2}{3} \right)^2 + \ldots \right) \] \[ = e \left( 1 + \left( -\frac{x}{2} + \frac{x^2}{3} \right) + \frac{1}{2} \left( \frac{x^2}{4} - \frac{2x^3}{6} + \frac{x^4}{9} \right) + \ldots \right) \] Thus, \[ y - e + \frac{1}{2} e^x = e \left[ \frac{1}{3} x^2 + \left(\frac{1}{2} \times \frac{x^2}{4}\right) \right] + \ldots = e \left[ \frac{1}{3} x^2 + \frac{1}{8} x^2 \right] + \ldots \] \[ = e \left[ \frac{3}{8} x^2 \right] + \ldots \] Therefore, \[ \lim_{x \to 0} \frac{y - e + \frac{1}{2} e^x}{x^2} = e \left[ \frac{1}{3} + \frac{1}{8} \right] = \frac{11}{24} e \]