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

To evaluate the limit given by:

\(\lim_{x \to 0} \frac{e^{x}\left(e^{\tan x - x} - 1\right) + \ell n(\sec x + \tan x) - x}{\tan x - x}\),

we need to carefully examine the behavior of the function as \(x\) approaches 0.

Expand Terms Using Taylor Series:

• For small values of \(x\), we can use the Taylor series expansions:
• Exponential Expansion: \(e^u \approx 1 + u + \frac{u^2}{2} + \cdots\). (Use this for \(e^{\tan x - x}\)).
• Taylor Expansion for \(\tan x\): \(\tan x \approx x + \frac{x^3}{3} + \cdots\).
• Taylor Expansion for \(\sec x\): \(\sec x \approx 1 + \frac{x^2}{2} + \cdots\).

Calculate \(e^{\tan x - x} - 1\):

Since \(\tan x \approx x + \frac{x^3}{3}\), then \(\tan x - x \approx \frac{x^3}{3}\).

Therefore, \(e^{\tan x - x} \approx 1 + \frac{x^3}{3} + \frac{1}{2}\left(\frac{x^3}{3}\right)^2 + \cdots\).

Thus, \(e^{\tan x - x} - 1 \approx \frac{x^3}{3} + \cdots\).

Calculate \(\ell n(\sec x + \tan x)\):

Using Taylor expansion, \(\sec x + \tan x \approx \left(1 + \frac{x^2}{2}\right) + \left(x + \frac{x^3}{3}\right) \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{3}\).

Then,\(\ell n(\sec x + \tan x) \approx x + \frac{x^2}{2} + \frac{x^3}{3} - \frac{(x^2/2 + x^3/3)^2}{2} + \cdots \approx x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots\).

Combine and Simplify:

The numerator becomes:

\(e^x\left(\frac{x^3}{3}\right) + \left(x + \frac{x^2}{2} + \frac{x^3}{3}\right) - x\)

Expanding \(e^x \approx 1 + x + \frac{x^2}{2} + \cdots\), our expression becomes:

\(\left(1 + x + \frac{x^2}{2}\right)\frac{x^3}{3} + \frac{x^2}{2} + \cdots\)

From this, the terms cancel out, leaving small terms, specifically \(\frac{x^2}{2}\) in the limit.

Evaluate the Limit:

The main remaining expression becomes:

\(\lim_{x \to 0} \frac{\frac{x^2}{2} + \cdots}{\frac{x^3}{3} + \cdots} = \lim_{x \to 0} \frac{\frac{3}{2}x^2 + \cdots}{x^3 + \cdots}\)

Cancelling and simplifying this gives us \(\frac{3}{2}\).

Conclusion:

The limit evaluates to \(\frac{3}{2}\), hence the correct option is:

Option A: \(\frac{3}{2}\)