Questions: Use L'Hôpital's Rule to evaluate lim x → ∞ (x^2 / e^(7x))

Solution Steps

To evaluate the limit \(\lim _{x \rightarrow \infty} \frac{x^{2}}{e^{7 x}}\) using L'Hôpital's Rule, we need to recognize that the limit is in the indeterminate form \(\frac{\infty}{\infty}\). L'Hôpital's Rule states that if the limit of \(\frac{f(x)}{g(x)}\) as \(x\) approaches a point results in an indeterminate form, then it can be evaluated as \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\). We will apply L'Hôpital's Rule repeatedly until we can evaluate the limit.

1. Identify the indeterminate form \(\frac{\infty}{\infty}\).
2. Apply L'Hôpital's Rule by differentiating the numerator and the denominator.
3. Repeat the process until the limit can be evaluated directly.

We need to evaluate the limit: \[ \lim_{x \rightarrow \infty} \frac{x^{2}}{e^{7x}} \] This limit is in the indeterminate form \(\frac{\infty}{\infty}\).

Since the limit is in the form \(\frac{\infty}{\infty}\), we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator:

• The derivative of the numerator \(f(x) = x^{2}\) is \(f'(x) = 2x\).
• The derivative of the denominator \(g(x) = e^{7x}\) is \(g'(x) = 7e^{7x}\).

Thus, we have: \[ \lim_{x \rightarrow \infty} \frac{x^{2}}{e^{7x}} = \lim_{x \rightarrow \infty} \frac{2x}{7e^{7x}} \]

The new limit \(\frac{2x}{7e^{7x}}\) is still in the form \(\frac{\infty}{\infty}\). We apply L'Hôpital's Rule again:

• The derivative of the numerator \(2x\) is \(2\).
• The derivative of the denominator \(7e^{7x}\) is \(49e^{7x}\).

Now we have: \[ \lim_{x \rightarrow \infty} \frac{2x}{7e^{7x}} = \lim_{x \rightarrow \infty} \frac{2}{49e^{7x}} \]

As \(x\) approaches infinity, \(e^{7x}\) approaches infinity, making the limit: \[ \lim_{x \rightarrow \infty} \frac{2}{49e^{7x}} = 0 \]

Final Answer