1. Write the expression in a useful form

We need to evaluate

$$L=\lim_{x\to\infty}\left(\frac{x^2+5x+3}{x^2+x+2}\right)^x.$$

First simplify the fraction inside:

$$\frac{x^2+5x+3}{x^2+x+2} =1+\frac{(x^2+5x+3)-(x^2+x+2)}{x^2+x+2} =1+\frac{4x+1}{x^2+x+2}.$$

So,

$$L=\lim_{x\to\infty}\left(1+\frac{4x+1}{x^2+x+2}\right)^x.$$

2. Identify the standard limit form

As $x\to\infty$,

$$\frac{4x+1}{x^2+x+2}\to 0.$$

Hence the expression is of the form

$$(1+u_x)^x, \quad u_x\to 0.$$

We use the standard result:

$$\lim_{x\to\infty}(1+u_x)^x=e^{\lim x u_x}$$

provided the limit $\lim x u_x$ exists.

Here,

$$u_x=\frac{4x+1}{x^2+x+2}.$$

Then

$$xu_x=x\cdot \frac{4x+1}{x^2+x+2} =\frac{4x^2+x}{x^2+x+2}.$$

Now taking limit,

$$\lim_{x\to\infty}\frac{4x^2+x}{x^2+x+2}=4.$$

Therefore,

$$L=e^4.$$

3. Check the options

• A: $e^4$ ✅
• B: $e^2$ ❌
• C: $e^3$ ❌
• D: $1$ ❌

So the correct option is A.