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Question
If f(2) = 4, f′(2) = 1 then find `lim_(x -> 2) [(x"f"(2) - 2"f"(x))/(x - 2)]`
Solution
Put x = 2 + h. Then as x → 2, h → 0.
Also, x – 2 = h
∴ `lim_(x -> 2) (x"f"(2) - 2"f"(x))/(x - 2)`
= `lim_("h" -> 0) ((2 + "h")(2) - 2"f"(2 + "h"))/"h"`
= `lim_("h" -> 0) ("hf"(2) - 2["f"(2 + "h") - "f"(2)])/"h"`
= `lim_("h" -> 0) [("hf"(2))/"h" - (2["f"(2 + "h") - "f"(2)])/"h"]`
= `lim_("h" -> 0) "f"(2) - 2 lim_("h" -> 0) ("f"(2 + "h") - "f"(2))/"h"` ...[∵ h → 0, ∴ h ≠ 0]
= f(2) – 2f'(2)
= 4 – 2(1) ...[∵ f(2) = 4, f'(2) = 1]
= 2
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