mathop {lim }limits_{x to 4} left[ {frac{{{x^ | vedclass

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(B) माना $y = \mathop {\lim }\limits_{x 	o 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} ight]$ है।सूत्र $\mathop {\lim }\limits_{x 	o a} \frac{{{x^

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) माना $y = \mathop {\lim }\limits_{x 	o 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} ight]$ है।सूत्र $\mathop {\lim }\limits_{x 	o a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$ का उपयोग करने पर:यहाँ $n = 3/2$ और $a = 4$ है।$y = \frac{3}{2} 	imes {(4)^{(3/2) - 1}} = \frac{3}{2} 	imes {(4)^{1/2}} = \frac{3}{2} 	imes 2 = 3$.वैकल्पिक रूप से,$L$-Hospital नियम लागू करने पर:$\mathop {\lim }\limits_{x 	o 4} \frac{{\frac{d}{{dx}}({x^{3/2}} - 8)}}{{\frac{d}{{dx}}(x - 4)}} = \mathop {\lim }\limits_{x 	o 4} \frac{{\frac{3}{2}{x^{1/2}}}}{1} = \frac{3}{2} 	imes \sqrt{4} = \frac{3}{2} 	imes 2 = 3$.

$\mathop {\lim }\limits_{x \to 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} \right] = $

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