▶️ Answer/Explanation
Solution
To apply the squeeze theorem, we need \( g(x) \leq f(x) \leq h(x) \) near \( x = a \), and \( \lim_{x \to a} g(x) = \lim_{x \to a} h(x) \).
The problem confirms this condition holds, with \( g \) and \( h \) converging to the same limit.
The graph in option D shows \( f \) squeezed between \( g \) and \( h \), both meeting at \( x = a \).
✅ Answer: D)
▶️ Answer/Explanation
Solution
To find \( \lim_{x \to 2} f(x) \), we evaluate the limits of \( g(x) \) and \( h(x) \) at \( x = 2 \):
\[ \lim_{x \to 2} g(x) = \frac{1}{2^{2} – 4(2) + 5} = \frac{1}{4 – 8 + 5} = \frac{1}{1} = 1 \]
\[ \lim_{x \to 2} h(x) = \frac{2(2)^{2} – 8(2) + 10}{2^{2} – 4(2) + 6} = \frac{8 – 16 + 10}{4 – 8 + 6} = \frac{2}{2} = 1 \]
Since \( g(x) \leq f(x) \leq h(x) \) and both \( g(x) \) and \( h(x) \) approach 1 as \( x \to 2 \), the Squeeze Theorem guarantees that:
\[ \lim_{x \to 2} f(x) = 1 \]
✅ Answer: B) 1
▶️ Answer/Explanation
Solution
The Squeeze Theorem states that if \( g(x) \leq f(x) \leq h(x) \) near \( x = a \), and \( \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \), then \( \lim_{x \to a} f(x) = L \).
Option D satisfies these conditions with \( L = 12 \), so it implies \( \lim_{x \to 5} f(x) = 12 \).
Options A and B are insufficient (only one-sided bounds), and Option C does not guarantee convergence to 12.
✅ Answer: D)
▶️ Answer/Explanation
Solution
For any real number \(\Theta\), \(-1 \leq \sin \Theta \leq 1\). Therefore:
\[ \left| \frac{\sin \Theta}{\Theta} \right| \leq \frac{1}{|\Theta|} \]
As \(\Theta \rightarrow -\infty\), \(\frac{1}{|\Theta|} \rightarrow 0\). By the Squeeze Theorem:
\[ \lim_{\Theta \rightarrow -\infty} \frac{\sin \Theta}{\Theta} = 0 \]
✅ Answer: A) 0