Limits Exam 64E889

1. **Problem Statement:** Calculate the limit $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$. 2. **Formula and Rules:** The limit of a function as $x$ approaches a value $a$ is the value that the function approaches as $x$ gets closer to $a$. If direct substitution results in an indeterminate form like $\frac{0}{0}$, we simplify the expression. 3. **Step-by-step Solution:** - Substitute $x=2$ directly: $$\frac{2^2 - 4}{2 - 2} = \frac{4 - 4}{0} = \frac{0}{0}$$ which is indeterminate. - Factor the numerator: $$\frac{(x - 2)(x + 2)}{x - 2}$$. - Cancel the common factor: $$\frac{\cancel{(x - 2)}(x + 2)}{\cancel{(x - 2)}} = x + 2$$. - Now substitute $x=2$: $$2 + 2 = 4$$. 4. **Answer:** $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$. --- 1. **Problem Statement:** Find $$\lim_{x \to 0} \frac{\sin x}{x}$$. 2. **Formula and Rules:** The standard limit $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ is a fundamental trigonometric limit. 3. **Answer:** $$1$$. --- 1. **Problem Statement:** Evaluate $$\lim_{x \to \infty} \frac{3x^2 + 5}{2x^2 - x}$$. 2. **Formula and Rules:** For limits at infinity involving rational functions, divide numerator and denominator by the highest power of $x$ in the denominator. 3. **Step-by-step Solution:** - Divide numerator and denominator by $x^2$: $$\frac{3 + \frac{5}{x^2}}{2 - \frac{1}{x}}$$. - As $x \to \infty$, terms with $\frac{1}{x}$ or $\frac{1}{x^2}$ approach 0. - So the limit is $$\frac{3 + 0}{2 - 0} = \frac{3}{2}$$. 4. **Answer:** $$\frac{3}{2}$$. --- 1. **Problem Statement:** Determine $$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}$$. 2. **Formula and Rules:** When direct substitution yields $\frac{0}{0}$, rationalize the numerator or denominator to simplify. 3. **Step-by-step Solution:** - Substitute $x=1$: $$\frac{\sqrt{1} - 1}{1 - 1} = \frac{0}{0}$$ indeterminate. - Multiply numerator and denominator by the conjugate of numerator: $$\frac{\sqrt{x} - 1}{x - 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \frac{x - 1}{(x - 1)(\sqrt{x} + 1)}$$. - Cancel $x - 1$: $$\frac{\cancel{x - 1}}{\cancel{x - 1}(\sqrt{x} + 1)} = \frac{1}{\sqrt{x} + 1}$$. - Substitute $x=1$: $$\frac{1}{\sqrt{1} + 1} = \frac{1}{2}$$. 4. **Answer:** $$\frac{1}{2}$$.