AP Calculus AB : 1.7 Selecting Procedures for Determining Limits- Exam Style questions with Answer- MCQ

Question

\(\lim_{x\rightarrow -7}\frac{x+7}{\left | x+7 \right |}\) is

A) −1
B) 0
C) 1
D) nonexistent

▶️ Answer/Explanation

Solution

Correct Answer: D

Analyze left-hand and right-hand limits:

Right-hand limit (x → -7⁺):

\(\lim_{x\rightarrow -7^+}\frac{x+7}{|x+7|} = \lim_{x\rightarrow -7^+}\frac{x+7}{x+7} = 1\)

Left-hand limit (x → -7⁻):

\(\lim_{x\rightarrow -7^-}\frac{x+7}{|x+7|} = \lim_{x\rightarrow -7^-}\frac{x+7}{-(x+7)} = -1\)

Conclusion:

Since the left-hand limit (−1) ≠ right-hand limit (1), the limit does not exist.

Question

If f is the function defined above, then \(\lim_{x \to 0}f(x)\) is

A) -1
B) 0
C) 1
D) nonexistent

▶️ Answer/Explanation

Solution

Correct Answer: D

Explanation:

The function shows that as x approaches 0 from the left (x→0⁻), the function approaches 1,

but as x approaches 0 from the right (x→0⁺), the function approaches -1.

Since these one-sided limits are not equal, the limit does not exist.

If f is the function defined by \(f(x)=\frac{x-1}{1-\frac{1}{x}}\), then \(\lim_{x \to 1}f(x)\) is equivalent to:

A)
B)
C)
D)

▶️ Answer/Explanation

Answer: A

Simplify \(f(x)\):
\(\frac{x-1}{1-\frac{1}{x}} = \frac{x-1}{\frac{x-1}{x}} = x\)

Thus, \(\lim_{x \to 1}f(x) = 1\)

Let f and g be functions such that \(\lim_{x \to 4}g(x)=7\) and \(\lim_{x \to 4}{\frac{f(x)}{g(x)}}=\pi\). What is \(\lim_{x \to 4}{f(x)}\)?

A) \(\frac{\pi}{7}\)
B) \(7 + \pi\)
C) \(7\pi\)
D) The limit cannot be determined

▶️ Answer/Explanation

Answer: C

Given:
\(\lim_{x \to 4}\frac{f(x)}{g(x)} = \pi\)
\(\lim_{x \to 4}g(x) = 7\)

Therefore:
\(\lim_{x \to 4}f(x) = \pi \times 7 = 7\pi\)

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