Question:

<p><strong>Step 1: Verify indeterminate form</strong></p><p>At x = 1: Numerator = f(1)g(1) - f(1)g(1) = 0, and Denominator = g(1) - f(1) = 2 - 2 = 0. This is 0/0 form.</p><p><strong>Step 2: Algebraic manipulation</strong></p><p>Add and subtract f(1)g(1) in the numerator:</p><p>f(1)g(x) - f(x)g(1) = f(1)g(x) - f(1)g(1) + f(1)g(1) - f(x)g(1)</p><p>= f(1)[g(x) - g(1)] - g(1)[f(x) - f(1)]</p><p><strong>Step 3: Rewrite the limit</strong></p><p>$$\lim_{x \to 1} \frac{f(1)[g(x) - g(1)] - g(1)[f(x) - f(1)]}{g(x) - f(x)}$$</p><p><strong>Step 4: Divide numerator and denominator by (x - 1)</strong></p><p>$$\lim_{x \to 1} \frac{f(1)\cdot\frac{g(x)-g(1)}{x-1} - g(1)\cdot\frac{f(x)-f(1)}{x-1}}{\frac{g(x)-f(x)}{x-1}}$$</p><p><strong>Step 5: Apply derivative definition</strong></p><p>$$= \frac{f(1)\cdot g'(1) - g(1)\cdot f'(1)}{g'(1) - f'(1)} = \frac{2g'(1) - 2f'(1)}{g'(1) - f'(1)}$$</p><p><strong>Step 6: Factor and simplify</strong></p><p>$$= \frac{2[g'(1) - f'(1)]}{g'(1) - f'(1)} = 2$$</p><p>∴ Answer: <strong>2</strong></p>