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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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Give step-by-step solution with explanation and final answer:Uploaded Image2. Calculate the given limits using L'Hopital’s rule: sin(a)—x meg. CR ES ME

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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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