▶️ Answer/Explanation
Solution
We analyze the limit from both sides of \( x = -7 \).
1. Left-hand limit:
If \( x < -7 \), then \( x + 7 < 0 \), so \( |x + 7| = -(x + 7) \)
\[ \lim_{x \to -7^-} \frac{x + 7}{|x + 7|} = \frac{x + 7}{-(x + 7)} = -1 \]
2. Right-hand limit:
If \( x > -7 \), then \( x + 7 > 0 \), so \( |x + 7| = x + 7 \)
\[ \lim_{x \to -7^+} \frac{x + 7}{|x + 7|} = \frac{x + 7}{x + 7} = 1 \]
Since the left-hand and right-hand limits are not equal, the limit does not exist.
✅ Answer: D) nonexistent
▶️ Answer/Explanation
Solution
We evaluate the limit by analyzing the behavior of \( f(x) = \frac{x}{|x|} \) as \( x \to 0 \) from both sides.
Left-hand limit:
As \( x \to 0^- \), \( x < 0 \), so \( |x| = -x \).
Thus, \( f(x) = \frac{x}{|x|} = \frac{x}{-x} \).
\[ \lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{-x} = \lim_{x \to 0^-} (-1) = -1 \]
Right-hand limit:
As \( x \to 0^+ \), \( x > 0 \), so \( |x| = x \).
Thus, \( f(x) = \frac{x}{|x|} = \frac{x}{x} \).
\[ \lim_{x \to 0^+} \frac{x}{|x|} = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} 1 = 1 \]
Since the left-hand limit (\( -1 \)) and right-hand limit (\( 1 \)) are not equal, the limit does not exist.
✅ Answer: D) nonexistent
▶️ Answer/Explanation
Solution
We simplify the function \( f(x) = \frac{x-1}{1-\frac{1}{x}} \) for \( x \neq 1 \).
Rewrite the denominator: \( 1 – \frac{1}{x} = \frac{x}{x} – \frac{1}{x} = \frac{x-1}{x} \).
Thus, \( f(x) = \frac{x-1}{\frac{x-1}{x}} = (x-1) \cdot \frac{x}{x-1} \).
The \( x-1 \) terms cancel out, leaving \( f(x) = x \) for \( x \neq 1 \).
Now evaluate the limit: \( \lim_{x \to 1} f(x) = \lim_{x \to 1} x \).
Since \( \lim_{x \to 1} x = 1 \), the limit is \( \lim_{x \to 1} x \).
✅ Answer: A) \( \lim_{x \to 1} x \)
▶️ Answer/Explanation
Solution
We need to find \( \lim_{x \to 4} f(x) \) given \( \lim_{x \to 4} \frac{f(x)}{g(x)} = \pi \) and \( \lim_{x \to 4} g(x) = 7 \).
Since \( \lim_{x \to 4} g(x) = 7 \neq 0 \), the quotient property for limits can be applied.
The quotient property states that if \( \lim_{x \to a} g(x) \neq 0 \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \).
Applying this property at \( x \to 4 \):
\[ \lim_{x \to 4} \frac{f(x)}{g(x)} = \frac{\lim_{x \to 4} f(x)}{\lim_{x \to 4} g(x)} \]
Given \( \lim_{x \to 4} \frac{f(x)}{g(x)} = \pi \) and \( \lim_{x \to 4} g(x) = 7 \), we have:
\[ \pi = \frac{\lim_{x \to 4} f(x)}{7} \]
Solving for \( \lim_{x \to 4} f(x) \):
\[ \lim_{x \to 4} f(x) = 7 \cdot \pi = 7\pi \]
Therefore, the limit is \( 7\pi \).
✅ Answer: C) \( 7\pi \)