Question

Evaluate \[ \lim_{n\to\infty}\frac{A+e^{nx}}{x+Ae^{nx}} \]

• A. \(\dfrac{A}{x}, \text{ when } x\lt 0\)
• B. \(1, \text{ when } x\gt 0\)
• C. \(0, \forall x\in\mathbb{R}\)
• D. \(A, \text{ when } x=0\)

Hint:

For limits involving \(e^{nx}\): \[ x\lt 0 \Rightarrow e^{nx}\to 0, \] and \[ x\gt 0 \Rightarrow e^{nx}\to \infty. \] Always analyse the limit separately for different signs of \(x\).

Answer 1

The Correct Option is A

Step 1: Consider the case \(x\lt 0\).
If \[ x\lt 0, \] then \[ nx\to -\infty \quad \text{as} \quad n\to\infty \] Hence, \[ e^{nx}\to 0 \] Therefore, \[ \frac{A+e^{nx}}{x+Ae^{nx}} \to \frac{A+0}{x+0} \] \[ = \frac{A}{x} \]

Step 2: Consider the case \(x\gt 0\).
If \[ x\gt 0, \] then \[ nx\to \infty \] Thus, \[ e^{nx}\to \infty \] Dividing numerator and denominator by \(e^{nx}\), \[ \frac{A+e^{nx}}{x+Ae^{nx}} = \frac{\frac{A}{e^{nx}}+1}{\frac{x}{e^{nx}}+A} \] As \[ n\to\infty, \] we get \[ \frac{0+1}{0+A} = \frac{1}{A} \] Hence option \((2)\) is not correct.

Step 3: Consider the case \(x=0\).
If \[ x=0, \] then \[ e^{nx}=1 \] Therefore, \[ \frac{A+1}{0+A} = \frac{A+1}{A} \] This is not equal to \(A\). Hence option \((4)\) is incorrect.

Step 4: Final conclusion.
Thus, the correct statement is \[ \boxed{\frac{A}{x}, \text{ when } x\lt 0} \]