Let [t] be the greatest integer less than or | vedclass

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

Question

(C) We know that $\lim_{x \rightarrow 0^{+}} x[\frac{k}{x}] = k$ for any constant $k > 0$. Applying this to the given limit: $\lim_{x \rightarrow 0^{+

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) We know that $\lim_{x ightarrow 0^{+}} x[\frac{k}{x}] = k$ for any constant $k > 0$. Applying this to the given limit: $\lim_{x ightarrow 0^{+}} x \sum_{k=1}^{p} [\frac{k}{x}] = \sum_{k=1}^{p} k = \frac{p(p+1)}{2}$. Similarly,$\lim_{x ightarrow 0^{+}} x^2 [\frac{k^2}{x^2}] = k^2$. So,$\lim_{x}$ ${ightarrow 0^{+}} x^2 \sum_{k=1}^{9} [\frac{k^2}{x^2}] = \sum_{k=1}^{9} k^2 = \frac{9(9+1)(2 	imes 9 + 1)}{6} = \frac{9 	imes 10 	imes 19}{6} = 285$. The inequality becomes: $\frac{p(p+1)}{2} - 285 \geq 1$ $\frac{p(p+1)}{2} \geq 286$ $p(p+1) \geq 572$. For $p=23$,$23 	imes 24 = 552 For $p=24$,$24 	imes 25 = 600 \geq 572$. Thus,the least natural value of $p$ is $24$.

Similar Questions

$\mathop {\lim }\limits_{x \to 0} {(\cos mx)^{n/{x^2}}}$ equals

$\mathop {Lim}\limits_{x \to 0} \frac{{\log _{{{\sin }^2}x}}\cos x}{{\log _{{{\sin }^2}\frac{x}{2}}}\cos \frac{x}{2}}$ has the value equal to

For $x \in R$,$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x}$ is equal to

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{ax+b}{cx+1}$

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module