Question

If \[ \lim_{x\to\infty} \left(\frac{ax^2+bx+c}{lx^2+mx+n}\right) \left(\frac{lx-1}{lx+a}\right)^{\frac{x}{2}} = \frac{3}{\sqrt{16e}}, \] then \[ \lim_{x\to0} \frac{l+mx+cx^2}{a+bx+nx^2} = \]

• A. \(\dfrac34\)
• B. \(\dfrac43\)
• C. \(\dfrac54\)
• D. \(\dfrac45\) \bigskip

Hint:

In limits of the form \[ \left(\frac{x+\alpha}{x+\beta}\right)^x, \] rewrite as \[ \left(1+\frac{\alpha-\beta}{x+\beta}\right)^x \] and use \[ \left(1+\frac{k}{x}\right)^x \to e^k. \]

Answer 1

The Correct Option is B

Concept: For large \(x\), \[ \frac{ax^2+bx+c}{lx^2+mx+n} \longrightarrow \frac{a}{l}. \] Also, \[ \left(1+\frac{k}{x}\right)^x \longrightarrow e^k. \]

Step 1: Evaluate the exponential limit. Consider \[ \left(\frac{lx-1}{lx+a}\right)^{x/2}. \] Write it as \[ \left( \frac{1-\frac1{lx}} {1+\frac{a}{lx}} \right)^{x/2}. \] Using \[ \ln\left(\frac{1-\frac1{lx}} {1+\frac{a}{lx}} \right) = -\frac{1+a}{lx}+o\!\left(\frac1x\right). \] Therefore, \[ \left(\frac{lx-1}{lx+a}\right)^{x/2} \to e^{-\frac{1+a}{2l}}. \] Given \[ \frac{3}{\sqrt{16e}} = \frac34\,e^{-1/2}. \] Hence \[ e^{-\frac{1+a}{2l}} = e^{-1/2}. \] Therefore, \[ \frac{1+a}{2l} = \frac12. \] \[ a+1=l. \] \[ \cdots (1) \]

Step 2: Compare the remaining constant factor. The rational factor tends to \[ \frac{a}{l}. \] Hence \[ \frac{a}{l} = \frac34. \] \[ 4a=3l. \] \[ \cdots (2) \] Using (1), \[ l=a+1. \] Substitute into (2): \[ 4a=3(a+1). \] \[ a=3. \] \[ l=4. \]

Step 3: Evaluate the required limit. \[ \lim_{x\to0} \frac{l+mx+cx^2} {a+bx+nx^2} = \frac{l}{a}. \] Substituting \[ l=4,\qquad a=3, \] \[ \lim_{x\to0} \frac{l+mx+cx^2} {a+bx+nx^2} = \frac43. \]

Step 4: Write the final answer. \[ \boxed{\frac43} \]