Limits Practice 11/11/24

Evaluate the limit without graphing.

• 0

• 1/4

• 6

• DNE

Answer explanation

To evaluate the limit, apply L'Hôpital's Rule or simplify the expression. The limit approaches 1/4 as the variable approaches the specified value, confirming that the correct answer is 1/4.

Find the limit of the function as x approaches 2⁺.

• 1

• -1

• 5

• DNE

Answer explanation

As x approaches 2 from the right, the function approaches 5. Therefore, the limit of the function as x approaches 2+ is 5.

Tags

CCSS.HSF.IF.A.2

• -1/4

• 8

• DNE

• ∞

• 3

• 6

• 9

• Does not exist

Answer explanation

To evaluate the limit, factor the numerator: \(x^2 - 9 = (x - 3)(x + 3)\). The expression simplifies to \(x + 3\) as \(x \to 3\). Thus, \(\lim_{x \to 3} (x + 3) = 6\). The correct answer is 6.

• 0

• Does not exist

Answer explanation

As x approaches 2 from the right (2+), the expression x - 2 becomes a small positive number. Thus, 1/(x - 2) approaches +∞. Therefore, the one-sided limit is +∞.

• 0

• 1

Answer explanation

To find the limit as x approaches infinity, divide the numerator and denominator by x^2: \lim_{x \to \infty} \frac{5 + \frac{3}{x} - \frac{2}{x^2}}{2 - \frac{1}{x} + \frac{1}{x^2}} = \frac{5}{2}. Thus, the correct answer is \frac{5}{2}.

• 0

• 1

• 5

• Does not exist

Answer explanation

Using the limit property, \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \) for any constant \( k \). Here, \( k = 5 \), so \( \lim_{x \to 0} \frac{\sin(5x)}{x} = 5 \). Thus, the limit is 5.

• 0

• 1

• 3

• Does not exist

Answer explanation

To evaluate \( \lim_{x \to -1} \frac{x^3 + 1}{x + 1} \), factor the numerator: \( x^3 + 1 = (x + 1)(x^2 - x + 1) \). The limit simplifies to \( \lim_{x \to -1} (x^2 - x + 1) = 3 \). Thus, the answer is 3.

• 1

• -1

• 0

• Does not exist

Answer explanation

As x approaches 0 from the left (0-), |x| equals -x. Thus, \( \frac{|x|}{x} = \frac{-x}{x} = -1 \). Therefore, the one-sided limit is -1.

• 0

• 1

• 3

Answer explanation

To find the limit as x approaches infinity, divide the numerator and denominator by x: \lim_{x \to \infty} \frac{3 + \frac{7}{x}}{1 - \frac{4}{x}}. As x approaches infinity, \frac{7}{x} \to 0 and \frac{4}{x} \to 0, resulting in \frac{3}{1} = 3.

Tags

CCSS.HSF-IF.C.7D