Precalculus Limit Laws: Practice Exam Questions

🤔 Precalculus Limit Laws: Practice Exam Questions

Let's test your knowledge of precalculus limit laws with some practice exam questions. These questions will help reinforce your understanding of how to apply these laws effectively.

Question 1: Basic Limit Evaluation

Evaluate the following limit: $$\lim_{x \to 2} (3x^2 - 2x + 1)$$ Solution: Using the limit laws, we can evaluate this limit by direct substitution: $$\lim_{x \to 2} (3x^2 - 2x + 1) = 3(2)^2 - 2(2) + 1$$ $$ = 3(4) - 4 + 1$$ $$ = 12 - 4 + 1$$ $$ = 9$$ Therefore, the limit is 9.

Question 2: Limit of a Quotient

Evaluate the following limit: $$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$$ Solution: First, notice that direct substitution results in an indeterminate form $\frac{0}{0}$. We can factor the numerator and simplify: $$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$$ $$ = \lim_{x \to 3} (x + 3)$$ Now, we can use direct substitution: $$\lim_{x \to 3} (x + 3) = 3 + 3 = 6$$ Thus, the limit is 6.

Question 3: Limit with a Radical

Evaluate the following limit: $$\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$$ Solution: Again, direct substitution leads to an indeterminate form. We can rationalize the numerator: $$\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \lim_{x \to 4} \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)}$$ $$ = \lim_{x \to 4} \frac{x - 4}{(x - 4)(\sqrt{x} + 2)}$$ $$ = \lim_{x \to 4} \frac{1}{\sqrt{x} + 2}$$ Now, we can use direct substitution: $$\lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$$ Therefore, the limit is $\frac{1}{4}$.

Question 4: Limit of a Piecewise Function

Consider the piecewise function: $$f(x) = \begin{cases} x^2, & x < 1 \\ 2x - 1, & x \geq 1 \end{cases}$$ Evaluate $\lim_{x \to 1} f(x)$. Solution: We need to check the left-hand limit and the right-hand limit. Left-hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = (1)^2 = 1$$ Right-hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1$$ Since the left-hand limit equals the right-hand limit, the limit exists and is equal to 1. $$\lim_{x \to 1} f(x) = 1$$

Question 5: Squeeze Theorem

Suppose we have $1 \leq f(x) \leq x^2 + 2x - 2$ for all $x$ near $1$. Find $\lim_{x \to 1} f(x)$. Solution: We can use the Squeeze Theorem. We need to find the limits of the bounding functions as $x$ approaches 1. $$\lim_{x \to 1} 1 = 1$$ $$\lim_{x \to 1} (x^2 + 2x - 2) = (1)^2 + 2(1) - 2 = 1 + 2 - 2 = 1$$ Since both limits are equal to 1, by the Squeeze Theorem: $$\lim_{x \to 1} f(x) = 1$$

🎉 Conclusion

These practice questions should give you a solid foundation in applying precalculus limit laws. Keep practicing, and you'll master these concepts in no time! 🚀