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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $y = \mathop {\lim }\limits_{x 	o 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} ight]$.Using the formula $\mathop {\lim }\limits_{x 	o a} \f

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $y = \mathop {\lim }\limits_{x 	o 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} ight]$.Using the formula $\mathop {\lim }\limits_{x 	o a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$:Here $n = 3/2$ and $a = 4$.$y = \frac{3}{2} 	imes {(4)^{(3/2) - 1}} = \frac{3}{2} 	imes {(4)^{1/2}} = \frac{3}{2} 	imes 2 = 3$.Alternatively,applying $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 4} \frac{{\frac{d}{{dx}}({x^{3/2}} - 8)}}{{\frac{d}{{dx}}(x - 4)}} = \mathop {\lim }\limits_{x 	o 4} \frac{{\frac{3}{2}{x^{1/2}}}}{1} = \frac{3}{2} 	imes \sqrt{4} = \frac{3}{2} 	imes 2 = 3$.

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

Similar Questions

Let $[.]$ denote the greatest integer function. Assertion $(A) : \lim_{x \rightarrow \infty} \frac{[x]}{x} = 1$. Reason $(R) : f(x) = x - 1, g(x) = [x], h(x) = x$ and $\lim_{x \rightarrow \infty} \frac{f(x)}{x} = \lim_{x \rightarrow \infty} \frac{h(x)}{x} = 1$.

Evaluate: $\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+x}-1}{x}$

$[x]$ denotes the greatest integer less than or equal to $x$. If $\{x\}=x-[x]$ and $\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$,then $\sin \theta+\cos \theta=$

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $

Evaluate the given limit: $\mathop {\lim }\limits_{z \to 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module