Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4.
Then lim _(x → a) (xf(a) - af(x))/(x - a) equals :

Solution

Maths sir · 4 steps

Add and subtract af(a) in the numerator, and the definition of f'(a) appears by itself.

1. Check the form. At x = a the numerator is af(a) - af(a) = 0 and the denominator is 0. So this is a 0/0 form, and we need to rewrite it.
2. Split the numerator. Subtract and add af(a): xf(a) - af(x) = (x - a)f(a) - a(f(x) - f(a)). Now divide each part by x - a.

   lim(x → a)(xf(a) - af(x))/(x - a) = f(a) - a lim(x → a)(f(x) - f(a))/(x - a) = f(a) - af'(a)

3. Cross-check. L'Hopital's rule gives the same result. Differentiate the top and bottom with respect to x to get (f(a) - af'(x))/1, which tends to f(a) - af'(a). Here f(a) is a fixed number, so its derivative with respect to x is just that number's coefficient, 1.
4. So. Put in f(a) = 4 and f'(a) = 2. The limit is 4 - 2a, which is option (a).

The idea to keep. When a 0/0 limit mixes f(x) and f(a), add and subtract a term so that (f(x) - f(a))/(x - a) appears. Remember that f(a) is a constant, not a function of x.

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