The value of lim_{x to 0} (frac{8}{x^8}) [1 - | vedclass

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

Question

(B) Let the expression be $L = \lim_{x 	o 0} \frac{8}{x^8} [1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4}]$.Fac

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(B) Let the expression be $L = \lim_{x 	o 0} \frac{8}{x^8} [1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4}]$.Factor the expression inside the bracket: $[(1 - \cos \frac{x^2}{2}) - \cos \frac{x^2}{4}(1 - \cos \frac{x^2}{2})] = (1 - \cos \frac{x^2}{2})(1 - \cos \frac{x^2}{4})$.Using the limit formula $\lim_{	heta 	o 0} (1 - \cos 	heta) = \frac{	heta^2}{2}$, we have:$1 - \cos \frac{x^2}{2} \approx \frac{1}{2} (\frac{x^2}{2})^2 = \frac{x^4}{8}$.$1 - \cos \frac{x^2}{4} \approx \frac{1}{2} (\frac{x^2}{4})^2 = \frac{x^4}{32}$.Substituting these into the limit:$L = \lim_{x 	o 0} \frac{8}{x^8} \cdot (\frac{x^4}{8}) \cdot (\frac{x^4}{32}) = \lim_{x 	o 0} \frac{8 \cdot x^8}{8 \cdot 32 \cdot x^8} = \frac{1}{32}$.

The value of $\lim_{x \to 0} (\frac{8}{x^8}) [1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cdot \cos \frac{x^2}{4}]$ is equal to:

Similar Questions

Let $f(x) = \frac{x - 1}{2x^2 - 7x + 5}$. Then:

If $\mathop {\lim }\limits_{x \to 2} \frac{{{x^n} - {2^n}}}{{x - 2}} = 80$,where $n$ is a positive integer,then $n = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 + x} - \sqrt {3 - x} }}{x} = $

The value of $\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}\,\, - \,\,{{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}}\,$ (where $n \in N$) is

The value of $\lim _{x}$ ${\rightarrow 0} 2\left(\frac{1-\cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \ldots \sqrt[10]{\cos 10 x}}{x^2}\right)$ is ............

Vedclass Test Series

Exam Paper Generator

Online Exam Module