Evaluate the limit: lim _{x} {rightarrow 0}le | vedclass

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

Question

(C) Let the given expression be $L = \lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3}

Vedclass · Accepted Answer

$\frac{1}{24}$

Vedclass · Answer

(C) Let the given expression be $L = \lim _{x}$ ${ightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}ight)ight)$.We can factor the expression inside the limit as:$L = \lim _{x ightarrow 0} \frac{4!}{x^8} (1 - \cos \frac{x^2}{3})(1 - \cos \frac{x^2}{4})$.Using the identity $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$,we get:$1 - \cos \frac{x^2}{3} = 2 \sin^2(\frac{x^2}{6})$ and $1 - \cos \frac{x^2}{4} = 2 \sin^2(\frac{x^2}{8})$.Substituting these into the limit:$L = 24 	imes \lim _{x ightarrow 0} \frac{1}{x^8} 	imes 2 \sin^2(\frac{x^2}{6}) 	imes 2 \sin^2(\frac{x^2}{8})$.$L = 24 	imes 4 	imes \lim _{x}$ ${ightarrow 0} \left(\frac{\sin(\frac{x^2}{6})}{\frac{x^2}{6}}ight)^2 	imes (\frac{x^2}{6})^2 	imes \left(\frac{\sin(\frac{x^2}{8})}{\frac{x^2}{8}}ight)^2 	imes (\frac{x^2}{8})^2 	imes \frac{1}{x^8}$.Since $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we have:$L = 96 	imes 1^2 	imes \frac{x^4}{36} 	imes 1^2 	imes \frac{x^4}{64} 	imes \frac{1}{x^8}$.$L = 96 	imes \frac{1}{36 	imes 64} = \frac{96}{2304} = \frac{1}{24}$.

Evaluate the limit: $\lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)$

Similar Questions

Let $x_n = (2^n + 3^n)^{\frac{1}{2n}}$ for all natural numbers $n$. Then,

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$,then the number of integral values of $a$ such that $\lim _{x \rightarrow a}([x^2]-[x]^2)$ does not exist is:

If $f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}$,then

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right-hand limit $\lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$ is equal to a nonzero real number is:

If $f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x > 3 \end{cases}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module