Question

If $f(a) = 2$, $f'(a) = 1$, $g(a) = -1$, $g'(a) = 2$, then as $x$ approaches $a$, $\frac{g(x)f(a)-g(a)f(x)}{(x-a)}$ approaches}

• A. 3
• B. 5
• C.
• D.

Hint:

Recognize expressions of the form $\lim_{x \to a} \frac{f(x)g(a) - f(a)g(x)}{x-a}$ as variations of the difference quotient. Applying L'Hopital's Rule (differentiating numerator and denominator) often simplifies such limits quickly. The result is typically $f'(a)g(a) - f(a)g'(a)$ or similar, depending on the exact form.

Answer 1

The Correct Option is A

Step 1: Identify the form of the limit. We are asked to evaluate the limit: \[L = \lim_{x \to a} \frac{g(x)f(a)-g(a)f(x)}{(x-a)}\] If we substitute $x=a$ into the expression, the numerator becomes $g(a)f(a) - g(a)f(a) = 0$, and the denominator becomes $a-a = 0$. This is an indeterminate form of type $\frac{0}{0}$, so we can apply L'Hopital's Rule.
Step 2: Apply L'Hopital's Rule. L'Hopital's Rule states that if $\lim_{x \to a} \frac{h(x)}{k(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{h(x)}{k(x)} = \lim_{x \to a} \frac{h'(x)}{k'(x)}$, provided the latter limit exists. Let $h(x) = g(x)f(a)-g(a)f(x)$ and $k(x) = x-a$. Then $h'(x) = g'(x)f(a) - g(a)f'(x)$ (since $f(a)$ and $g(a)$ are constants). And $k'(x) = 1$. So, the limit becomes: \[L = \lim_{x \to a} \frac{g'(x)f(a) - g(a)f'(x)}{1}\] \[L = g'(a)f(a) - g(a)f'(a)\]
Step 3: Substitute the given values. We are given: $f(a) = 2$ $f'(a) = 1$ $g(a) = -1$ $g'(a) = 2$ Substitute these values into the expression for $L$: \[L = (2)(2) - (-1)(1)\] \[L = 4 - (-1)\] \[L = 4 + 1\] \[L = 5\]