▶️ Answer/Explanation
Solution
1. Direct Substitution:
\[ \frac{\cos(0) + 4e^0}{5e^0} = \frac{1 + 4(1)}{5(1)} = \frac{5}{5} = 1 \]
Since we don’t get an indeterminate form, direct substitution works.
2. Step-by-step Evaluation:
– \(\cos(0) = 1\)
– \(e^0 = 1\) for both numerator and denominator
– Numerator becomes \(1 + 4(1) = 5\)
– Denominator becomes \(5(1) = 5\)
– Final evaluation: \(\frac{5}{5} = 1\)
3. Verification:
For very small x (approaching 0), both \(\cos x\) and \(e^x\) approach 1, making the limit clearly 1.
✅ Answer: C) 1
▶️ Answer/Explanation
Solution
1. Understanding the Limit:
We need to find the left-hand limit as \( x \) approaches 1, denoted by \(\lim_{x \to 1^-} f(x)\).
2. Selecting the Correct Piece:
For \( x \to 1^- \) (approaching 1 from the left), we use the definition where \( x < 1 \):
\[ f(x) = 2x + 2 \]
3. Evaluating the Limit:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x + 2) = 2(1) + 2 = 4 \]
4. Verification:
– The left-hand limit uses the linear function \( 2x + 2 \)
– Direct substitution gives the exact value as \( x \) approaches 1
– The quadratic piece (\( x^2 + 4 \)) is irrelevant for this left-hand limit
✅ Answer: B) 4
▶️ Answer/Explanation
Solution
1. Find \(\lim_{x \to 4} f(x)\):
From the graph, as \( x \) approaches 4, \( f(x) \) approaches 3.
\[ \lim_{x \to 4} f(x) = 3 \]
2. Find \(\lim_{x \to 4} g(x)\):
From the graph, as \( x \) approaches 4, \( g(x) \) approaches 5.
\[ \lim_{x \to 4} g(x) = 5 \]
3. Evaluate the limit:
\[ \lim_{x \to 4} \frac{f(x)+6}{g(x)} = \frac{\lim_{x \to 4} f(x) + 6}{\lim_{x \to 4} g(x)} = \frac{3 + 6}{5} = \frac{9}{5} \]
4. Verification:
– The limit exists because \( g(x) \) approaches 5 (not 0)
– Both functions appear continuous at \( x = 4 \) from the graph
✅ Answer: C) \(\frac{9}{5}\)
▶️ Answer/Explanation
Solution
1. Recognize the Limit Definition:
This limit represents the derivative of \( f(x) = 8x^8 \) at \( x = \frac{1}{2} \):
\[ f’\left(\frac{1}{2}\right) = \lim_{h\rightarrow 0}\frac{8\left(\frac{1}{2}+h\right)^8 – 8\left(\frac{1}{2}\right)^8}{h} \]
2. Compute the Derivative:
First find the general derivative: \[ f'(x) = \frac{d}{dx}(8x^8) = 64x^7 \]
Then evaluate at \( x = \frac{1}{2} \): \[ f’\left(\frac{1}{2}\right) = 64\left(\frac{1}{2}\right)^7 = 64 \times \frac{1}{128} = \frac{1}{2} \]
3. Verification:
– The limit exists as it matches the derivative definition
– The function \( f(x) = 8x^8 \) is differentiable everywhere
– Direct computation confirms the result
✅ Answer: B) \(\frac{1}{2}\)