MCQ
$\mathop {\lim }\limits_{x \to 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} \right] = $
- A$3/2$
- ✓$3$
- C$2/3$
- D$1/3$
==> $y = \mathop {\lim }\limits_{x \to 4} \frac{{({x^{1/2}} - 2)(x + 4 + 2\sqrt x )}}{{(\sqrt x - 2)(\sqrt x + 2)}}$
==> $y = \mathop {\lim }\limits_{x \to 4} \frac{{(x + 4 + 2\sqrt x )}}{{(\sqrt x + 2)}}$$ = \frac{{4 + 4 + 2\sqrt 4 }}{{\sqrt 4 + 2}}$$ = \frac{{12}}{4} = 3$.
Trick : Applying $ L-$ Hospital’s rule, we get
$\mathop {\lim }\limits_{x \to 4} \frac{{\frac{3}{2}{x^{1/2}}}}{1}$$ = \frac{3}{2}{(4)^{1/2}} = 3.$
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વિધાન $2$:$\;\mathop \sum \limits_{r = 0}^n \left( {r + 1} \right)\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right){x^r}\; = {\left( {1 + x} \right)^n} + nx{\left( {1 + x} \right)^{n - 1}}$