We are asked to find the limit of the function $\frac{2x^2 + 3x + 4}{x-1}$ as $x$ approaches 1 from the right. We want to evaluate $\lim_{x \to 1^+} \frac{2x^2 + 3x + 4}{x-1}$.

1. Problem Description

We are asked to find the limit of the function $\frac{2x^2 + 3x + 4}{x-1}$ as $x$ approaches 1 from the right. We want to evaluate $\lim_{x \to 1^+} \frac{2x^2 + 3x + 4}{x-1}$.

2. Solution Steps

First, let's check the limit of the numerator and the denominator separately as $x$ approaches 1 from the right.

The numerator approaches $2(1)^2 + 3(1) + 4 = 2 + 3 + 4 = 9$.

The denominator approaches $1-1 = 0$.

Since the denominator approaches 0 and the numerator approaches 9, we need to investigate further.

As $x$ approaches 1 from the right, $x > 1$, so $x-1 > 0$. Therefore, $x-1$ approaches 0 from the positive side.

Thus, we have a constant number (9) divided by a number approaching 0 from the positive side.

This means the limit is positive infinity.

$\lim_{x \to 1^+} (2x^2 + 3x + 4) = 2(1)^2 + 3(1) + 4 = 9$

$\lim_{x \to 1^+} (x-1) = 0^+$

$\lim_{x \to 1^+} \frac{2x^2 + 3x + 4}{x-1} = \frac{9}{0^+} = +\infty$

3. Final Answer

The limit is $\infty$.