We need to find the limit of the expression $\sqrt{3x^2+7x+1}-\sqrt{3}x$ as $x$ approaches infinity.

1. Problem Description

We need to find the limit of the expression $\sqrt{3x^2+7x+1}-\sqrt{3}x$ as $x$ approaches infinity.

2. Solution Steps

The given limit is of the form $\infty - \infty$, which is an indeterminate form. To evaluate this limit, we can multiply and divide by the conjugate of the expression.

$$\lim_{x\to\infty} (\sqrt{3x^2+7x+1}-\sqrt{3}x)$$

Multiply and divide by the conjugate $\sqrt{3x^2+7x+1}+\sqrt{3}x$:

$$\lim_{x\to\infty} \frac{(\sqrt{3x^2+7x+1}-\sqrt{3}x)(\sqrt{3x^2+7x+1}+\sqrt{3}x)}{\sqrt{3x^2+7x+1}+\sqrt{3}x}$$

$$\lim_{x\to\infty} \frac{(3x^2+7x+1) - (\sqrt{3}x)^2}{\sqrt{3x^2+7x+1}+\sqrt{3}x}$$

$$\lim_{x\to\infty} \frac{3x^2+7x+1 - 3x^2}{\sqrt{3x^2+7x+1}+\sqrt{3}x}$$

$$\lim_{x\to\infty} \frac{7x+1}{\sqrt{3x^2+7x+1}+\sqrt{3}x}$$

Now, divide both the numerator and denominator by $x$:

$$\lim_{x\to\infty} \frac{7+\frac{1}{x}}{\sqrt{3+\frac{7}{x}+\frac{1}{x^2}}+\sqrt{3}}$$

As $x \to \infty$, $\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$. Therefore, the limit becomes:

$$\frac{7}{\sqrt{3}+\sqrt{3}} = \frac{7}{2\sqrt{3}}$$

Multiply the numerator and denominator by $\sqrt{3}$:

$$\frac{7\sqrt{3}}{2\sqrt{3}\sqrt{3}} = \frac{7\sqrt{3}}{2(3)} = \frac{7\sqrt{3}}{6}$$

3. Final Answer

The final answer is $\frac{7\sqrt{3}}{6}$