To solve this limit using L'Hôpital's Rule, we need to evaluate:

$\lim_{x \to \infty} \frac{5x^2 - 3x}{7x^2 + 1}$

Step 1: Check the Indeterminate Form

As $x \to \infty$:

• The numerator $5x^2 - 3x$ grows to infinity.
• The denominator $7x^2 + 1$ also grows to infinity.

Thus, we have an $\infty / \infty$ form, which qualifies for L'Hôpital's Rule.

Step 2: Differentiate the Numerator and Denominator

L'Hôpital's Rule states that if we have an indeterminate form of $\frac{\infty}{\infty}$, we can take the derivative of the numerator and the denominator:

1. Differentiate the numerator: $\frac{d}{dx}(5x^2 - 3x) = 10x - 3$.
2. Differentiate the denominator: $\frac{d}{dx}(7x^2 + 1) = 14x$.

So the limit becomes:

$\lim_{x \to \infty} \frac{10x - 3}{14x}$

Step 3: Simplify the New Expression

Now, divide the terms in the fraction:

$\lim_{x \to \infty} \frac{10x - 3}{14x} = \lim_{x \to \infty} \frac{10x}{14x} - \frac{3}{14x} = \lim_{x \to \infty} \frac{10}{14} - \lim_{x \to \infty} \frac{3}{14x}$

As $x \to \infty$:

• The term $\frac{3}{14x} \to 0$.

Thus, we are left with:

$\frac{10}{14} = \frac{5}{7}$

Final Answer

$\lim_{x \to \infty} \frac{5x^2 - 3x}{7x^2 + 1} = \frac{5}{7}$

Would you like further details on any step, or do you have any additional questions?

Related Questions:

1. How would the result change if the coefficients in the polynomial were different?
2. Can L'Hôpital’s Rule be applied to functions other than polynomials?
3. What are the conditions for using L'Hôpital’s Rule?
4. What if the highest power terms in the numerator and denominator were different?
5. How does L'Hôpital's Rule handle indeterminate forms other than $\frac{\infty}{\infty}$?

Tip:

When dealing with limits at infinity for rational functions, check if the leading terms alone can help find the answer before applying L'Hôpital's Rule.