[x] represents the greatest integer function. | vedclass

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

Question

(A) Given the limit $\lim _{x \rightarrow 0^{+}} \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$. Since $x \rightarrow 0^{+}$,we have $[x] = 0$. Substituting t

Vedclass · Accepted Answer

$\sqrt{10}$

Vedclass · Answer

(A) Given the limit $\lim _{x ightarrow 0^{+}} \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$. Since $x ightarrow 0^{+}$,we have $[x] = 0$. Substituting this into the expression: $\lim _{x ightarrow 0^{+}} \frac{\cos(0) - \cos(kx - 0)}{x^2} = 5$ $\lim _{x ightarrow 0^{+}} \frac{1 - \cos(kx)}{x^2} = 5$ Using the limit formula $\lim _{	heta ightarrow 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$,we rewrite the expression: $\lim _{x ightarrow 0^{+}} \frac{1 - \cos(kx)}{(kx)^2} \cdot k^2 = 5$ $\frac{1}{2} \cdot k^2 = 5$ $k^2 = 10$ $k = \sqrt{10}$.

$[x]$ represents the greatest integer function. If $\lim _{x \rightarrow 0^{+}} \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$,then $k=$

Similar Questions

If $\lim _{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17$,then $5 a^2+b^2$ is equal to

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$,then:

If $\lim _{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}$ exists and is equal to $b$,then the value of $a-2b$ is ....... .

Define $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$. If $\lim_{x \rightarrow 0} f(x)$ exists,then $p =$

If $a_1 = 1$ and $a_{n+1} = \frac{4 + 3a_n}{3 + 2a_n}$ for $n \ge 1$,and if $\lim_{n \to \infty} a_n = a$,then the value of $a$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module