Give step-by-step solution with explanation and final answer:2. Calculate the given limits using L'Hopital’s rule: sin(a)—x meg. CR ES ME
Question:
Give step-by-step solution with explanation and final answer:
2. Calculate the given limits using L'Hopital’s rule:
sin(a)—x
meg.
CR ES ME
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Asked by: Test User 100251
Created at: 2025-05-21 08:18:38
Asked by: Test User 100251
Created at: 2025-05-21 08:18:38
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Let's solve both limits using **L'Hopital's Rule** step by step.
---
### 1. \(\lim_{x \to 0} \frac{\sin(x) - x}{x^2}\)
#### **Step 1:** Check the form
As \(x \to 0\):
- \(\sin(x) \to 0\)
- \(x \to 0\)
- Numerator: \(\sin(x) - x \to 0\)
- Denominator: \(x^2 \to 0\)
So, the limit is of the indeterminate form \(\frac{0}{0}\).
#### **Step 2:** Apply L'Hopital's Rule
Take the derivative of the numerator and denominator:
- Numerator's derivative: \(\cos(x) - 1\)
- Denominator's derivative: \(2x\)
So, the limit becomes:
\[
\lim_{x \to 0} \frac{\cos(x) - 1}{2x}
\]
#### **Step 3:** Check the form again
As \(x \to 0\):
- \(\cos(0) - 1 = 0\)
- \(2x \to 0\)
Still \(\frac{0}{0}\) form. Apply L'Hopital's Rule again.
#### **Step 4:** Second Application of L'Hopital's Rule
Take derivatives again:
- Numerator's derivative: \(-\sin(x)\)
- Denominator's derivative: \(2\)
So, the limit now is:
\[
\lim_{x \to 0} \frac{-\sin(x)}{2}
\]
#### **Step 5:** Substitute \(x = 0\)
- \(-\sin(0) = 0\)
So, the limit is:
\[
\boxed{0}
\]
---
### 2. \(\lim_{x \to \infty} x^2 e^{-x}\)
#### **Step 1:** Rewrite for L'Hopital
As \(x \to \infty\):
- \(x^2 \to \infty\)
- \(e^{-x} \to 0\)
So, the limit is of the form \(\infty \cdot 0\), which is indeterminate.
Rewrite as:
\[
x^2 e^{-x} = \frac{x^2}{e^{x}}
\]
Now, as \(x \to \infty\), this is \(\frac{\infty}{\infty}\) form.
#### **Step 2:** Apply L'Hopital's Rule
Take derivatives:
- Numerator's derivative: \(2x\)
- Denominator's derivative: \(e^{x}\)
So, the limit becomes:
\[
\lim_{x \to \infty} \frac{2x}{e^{x}}
\]
Still \(\frac{\infty}{\infty}\), apply L'Hopital's Rule again.
#### **Step 3:** Second Application
- Numerator's derivative: \(2\)
- Denominator's derivative: \(e^{x}\)
So, the limit is:
\[
\lim_{x \to \infty} \frac{2}{e^{x}}
\]
As \(x \to \infty\), \(e^{x} \to \infty\), so \(\frac{2}{e^{x}} \to 0\).
#### **Final Answer:**
\[
\boxed{0}
\]
---
## **Final Answers:**
1. \(\displaystyle \lim_{x \to 0} \frac{\sin(x) - x}{x^2} = \boxed{0}\)
2. \(\displaystyle \lim_{x \to \infty} x^2 e^{-x} = \boxed{0}\)
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