The problem asks us to evaluate three limits, $A$, $B$, and $C$. $A = \lim_{x\to1} \frac{1+x}{\sqrt{x^2+3}-2}$ $B = \lim_{x\to+\infty} \frac{x^2 + \ln x}{3x^2 - 1}$ $C = \lim_{x\to+\infty} \frac{x^3 - e^x}{2e^x + 3x + 1}$

1. Problem Description

The problem asks us to evaluate three limits, $A$, $B$, and $C$.

$A = \lim_{x\to1} \frac{1+x}{\sqrt{x^2+3}-2}$

$B = \lim_{x\to+\infty} \frac{x^2 + \ln x}{3x^2 - 1}$

$C = \lim_{x\to+\infty} \frac{x^3 - e^x}{2e^x + 3x + 1}$

2. Solution Steps

First, we evaluate $A$:

$A = \lim_{x\to1} \frac{1+x}{\sqrt{x^2+3}-2}$

When we substitute $x=1$ we get $\frac{1+1}{\sqrt{1^2+3}-2} = \frac{2}{\sqrt{4}-2} = \frac{2}{2-2} = \frac{2}{0}$. This is an indeterminate form. We can rationalize the denominator.

$A = \lim_{x\to1} \frac{1+x}{\sqrt{x^2+3}-2} \times \frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2} = \lim_{x\to1} \frac{(1+x)(\sqrt{x^2+3}+2)}{x^2+3-4} = \lim_{x\to1} \frac{(1+x)(\sqrt{x^2+3}+2)}{x^2-1} = \lim_{x\to1} \frac{(1+x)(\sqrt{x^2+3}+2)}{(x-1)(x+1)} = \lim_{x\to1} \frac{\sqrt{x^2+3}+2}{x-1}$

When we substitute $x=1$ we get $\frac{\sqrt{1^2+3}+2}{1-1} = \frac{\sqrt{4}+2}{0} = \frac{4}{0}$.

Since we approach $1$ from both directions, we either approach $\infty$ or $-\infty$.

For $x > 1$, the expression is positive, so the limit is $+\infty$.

Next, we evaluate $B$:

$B = \lim_{x\to+\infty} \frac{x^2 + \ln x}{3x^2 - 1}$

We can divide both numerator and denominator by $x^2$.

$B = \lim_{x\to+\infty} \frac{1 + \frac{\ln x}{x^2}}{3 - \frac{1}{x^2}}$

Since $\lim_{x\to\infty} \frac{\ln x}{x^2} = 0$ and $\lim_{x\to\infty} \frac{1}{x^2} = 0$, we have

$B = \frac{1 + 0}{3 - 0} = \frac{1}{3}$

Finally, we evaluate $C$:

$C = \lim_{x\to+\infty} \frac{x^3 - e^x}{2e^x + 3x + 1}$

We can divide both numerator and denominator by $e^x$.

$C = \lim_{x\to+\infty} \frac{\frac{x^3}{e^x} - 1}{2 + \frac{3x}{e^x} + \frac{1}{e^x}}$

Since $\lim_{x\to\infty} \frac{x^3}{e^x} = 0$, $\lim_{x\to\infty} \frac{3x}{e^x} = 0$ and $\lim_{x\to\infty} \frac{1}{e^x} = 0$, we have

$C = \frac{0 - 1}{2 + 0 + 0} = \frac{-1}{2} = -\frac{1}{2}$

3. Final Answer

$A = +\infty$

$B = \frac{1}{3}$

$C = -\frac{1}{2}$