MathematicsIf ⨍ (2) =4 and ⨍ 1(2)=1 then Ltx→2x⨍ (2)-2⨍ (x)x-2=

If $⨍ (2) =4$ and $⨍^{1} (2)=1$ then $\underset{x \to 2}{Lt} \frac{x⨍ (2)-2⨍ (x)}{x - 2} =$

Solution:

The given limit is $\underset{x \to 2}{Lt} \frac{xf(2)-2 f(x)}{x - 2}$

Use LH rule to get the above limit

$\begin{array}{rcl} \underset{x \to 2}{Lt} \frac{x⨍ (2)-2⨍ (x)}{x - 2} & = & \underset{x \to 2}{Lt} \frac{⨍ (2)-2⨍' (x)}{1} \\ = & f (2) - 2 f' (2) \\ = & 4 - 2 (1) \\ = & 2 \end{array}$

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