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, exponential limit: \(\left(1+\frac{k}{x}\right)^x \to e^k\). Manipulate the given limit to match this form, equate exponent and coefficient, then evaluate the small x limit.

Step 1: \(\left(\frac{lx-1}{lx+a}\right)^{x/2} = \left(\frac{1-1/(lx)}{1+a/(lx)}\right)^{x/2} \to e^{-(1+a)/(2l)}\). Given limit = \(\frac{3}{4}e^{-1/2}\), so \(\frac{1+a}{2l} = \frac12 \Rightarrow 1+a = l\).

Step 2: The rational part \(\frac{ax^2+bx+c}{lx^2+mx+n} \to \frac{a}{l} = \frac34 \Rightarrow 4a=3l\). With \(l=a+1\): \(4a=3a+3 \Rightarrow a=3, l=4\).

Step 3: \(\lim_{x\to0} \frac{l+mx+cx^2}{a+bx+nx^2} = \frac{l}{a} = \frac43\).

Step 4: Write the final answer. \(\boxed{\frac43}\)