What is the limit of the following, using L’Hôpital’s rule?

Ask by Ruiz Fuentes. in the United States

Dec 28,2024

Question

What is the limit of the following, using L'Hôpital's rule? $$ \lim _{x \rightarrow \infty} \frac{x^{3}-1}{e^{x}+1} $$

UpStudy AI · Accepted Answer

To evaluate the limit
$$
\lim_{x \rightarrow \infty} \frac{x^{3} - 1}{e^{x} + 1},
$$
we can use **L'Hôpital's Rule**, which is applicable when the limit is in an indeterminate form like $$\frac{\infty}{\infty}$$.

### Step-by-Step Application of L'Hôpital's Rule:

1. **Initial Form:**
   $$
   \lim_{x \rightarrow \infty} \frac{x^{3} - 1}{e^{x} + 1}
   $$
   As $$x \rightarrow \infty$$, both the numerator $$x^3 - 1$$ and the denominator $$e^x + 1$$ approach infinity. Thus, the form is $$\frac{\infty}{\infty}$$, and L'Hôpital's Rule can be applied.

2. **First Derivative:**
   $$
   \lim_{x \rightarrow \infty} \frac{3x^{2}}{e^{x}}
   $$
   The new form is still $$\frac{\infty}{\infty}$$, so we apply L'Hôpital's Rule again.

3. **Second Derivative:**
   $$
   \lim_{x \rightarrow \infty} \frac{6x}{e^{x}}
   $$
   Again, the form remains $$\frac{\infty}{\infty}$$, prompting another application of L'Hôpital's Rule.

4. **Third Derivative:**
   $$
   \lim_{x \rightarrow \infty} \frac{6}{e^{x}}
   $$
   Now, as $$x \rightarrow \infty$$, $$e^x$$ grows exponentially while the numerator remains constant. Therefore, the limit simplifies to:
   $$
   \lim_{x \rightarrow \infty} \frac{6}{e^{x}} = 0
   $$

### Conclusion:
Using L'Hôpital's Rule three times, we've determined that:
$$
\lim_{x \rightarrow \infty} \frac{x^{3} - 1}{e^{x} + 1} = 0
$$

**Answer:** 0
The limit is 0.

What is the limit of the following, using L’Hôpital’s rule?

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