Answered

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A. 0
B. 2

If [tex]f(x) = x^2 + 2 \ln x[/tex], then what is [tex]\lim_{A \rightarrow 0} \frac{f(2 + A) - f(2)}{A}[/tex]?

A. 5
B. 1
C. 2
D. 0

Sagot :

GinnyAnswer
To find [tex]\(\lim_{A \rightarrow 0} \frac{f(2 + A) - f(2)}{A}\)[/tex] for the function [tex]\(f(x) = x^2 + 2 \ln x\)[/tex], we can use the concept of the derivative. The expression given is the definition of the derivative of the function [tex]\(f(x)\)[/tex] at the point [tex]\(x=2\)[/tex].

To solve this step-by-step:

1. Identify the function [tex]\(f(x)\)[/tex] and the point of evaluation:
[tex]\[ f(x) = x^2 + 2 \ln x \][/tex]
The point of evaluation is [tex]\(x = 2\)[/tex].

2. Calculate the derivative of [tex]\(f(x)\)[/tex]:

The derivative [tex]\(f'(x)\)[/tex] of [tex]\(f(x)\)[/tex] is obtained by differentiating each term of [tex]\(f(x)\)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} (x^2) + \frac{d}{dx} (2 \ln x) \][/tex]
[tex]\[ f'(x) = 2x + \frac{2}{x} \][/tex]

3. Evaluate the derivative at [tex]\(x = 2\)[/tex]:
[tex]\[ f'(2) = 2(2) + \frac{2}{2} \][/tex]
[tex]\[ f'(2) = 4 + 1 \][/tex]
[tex]\[ f'(2) = 5 \][/tex]

4. Interpret the limit:

Using the definition of the derivative, we have:
[tex]\[ \lim_{A \rightarrow 0} \frac{f(2 + A) - f(2)}{A} = f'(2) \][/tex]

5. Conclusion:
[tex]\[ \lim_{A \rightarrow 0} \frac{f(2 + A) - f(2)}{A} = 5 \][/tex]
The correct answer is:
A. 5
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