Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{e^{x}\left(e^{\tan x - x} - 1\right) + \ell n(\sec x + \tan x) - x}{\tan x - x} \]

• A. $\dfrac{3}{2}$
• B. $\dfrac{3}{2}e$
• C. $\dfrac{5}{2}e$
• D. $\dfrac{5}{2}$

Hint:

For limits involving trigonometric and exponential functions near zero, always use Taylor series expansions up to the required order.

Answer 1

The Correct Option is A

Step 1: Use standard series expansions near $x=0$.
\[ \tan x = x + \frac{x^3}{3} + o(x^3), \quad \sec x = 1 + \frac{x^2}{2} + o(x^2) \] \[ \ell n(\sec x + \tan x) = x + \frac{x^3}{6} + o(x^3) \] Step 2: Expand each term in the numerator.
\[ e^{\tan x - x} - 1 = \frac{x^3}{3} + o(x^3) \] \[ e^x = 1 + x + o(x) \] \[ e^x\left(e^{\tan x - x} - 1\right) = \frac{x^3}{3} + o(x^3) \] Step 3: Simplify the expression.
\[ \text{Numerator} = \frac{x^3}{3} + \frac{x^3}{6} + o(x^3) = \frac{x^3}{2} + o(x^3) \] \[ \text{Denominator} = \tan x - x = \frac{x^3}{3} + o(x^3) \] Step 4: Evaluate the limit.
\[ \lim_{x \to 0} \frac{\frac{x^3}{2}}{\frac{x^3}{3}} = \frac{3}{2} \]