Let f : R → R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of lim _(x → 2) (x²f(2) - 4f(x))/(x - 2) is equal to :

Solution

Maths sir · 4 steps

Add and subtract 4f(2) and the limit splits into two derivatives you already know.

1. Check the form. Since f(2) = 4, the numerator is 4x² - 4f(x). At x = 2 it becomes 16 - 16 = 0, and the denominator is also 0. So this is a 0/0 form.
2. Split the numerator. Write 4x² - 4f(x) = 4(x² - 4) - 4(f(x) - 4). Here 4 = f(2), so the second bracket is f(x) - f(2).

   lim(x → 2)[4·(x² - 4)/(x - 2) - 4·(f(x) - f(2))/(x - 2)] = 4(2 + 2) - 4f'(2)

3. Use each piece. The first fraction is x + 2, which tends to 4, so that term gives 4 × 4 = 16. The second fraction is the definition of f'(2) = 1, so that term gives 4 × 1 = 4.
4. So. The limit is 16 - 4 = 12, option (d).

The idea to keep. When a limit has f(x) and x → a in a 0/0 form, rewrite it so that (f(x) - f(a))/(x - a) appears. That piece is just f'(a).

Answer from the NTA final answer key. Solution by PrepWiser.