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

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

Question

(B) Let the given expression be $L = \lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt

Vedclass · Accepted Answer

is equal to $27$

Vedclass · Answer

(B) Let the given expression be $L = \lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$. As $x \rightarrow \infty$,we can factor out $x$ from the terms inside the parentheses: Numerator: $(\sqrt{3x+1} \pm \sqrt{3x-1})^6 = x^3 (\sqrt{3+1/x} \pm \sqrt{3-1/x})^6$. Denominator: $(x \pm \sqrt{x^2-1})^6 = x^6 (1 \pm \sqrt{1-1/x^2})^6$. Substituting these into the limit: $L = \lim _{x}$ ${\rightarrow \infty} x^3 \cdot \frac{x^3 [(\sqrt{3+1/x} + \sqrt{3-1/x})^6 + (\sqrt{3+1/x} - \sqrt{3-1/x})^6]}{x^6 [(1 + \sqrt{1-1/x^2})^6 + (1 - \sqrt{1-1/x^2})^6]}$. As $x \rightarrow \infty$,$1/x \rightarrow 0$ and $1/x^2 \rightarrow 0$. $L = \frac{(\sqrt{3} + \sqrt{3})^6 + (\sqrt{3} - \sqrt{3})^6}{(1 + 1)^6 + (1 - 1)^6} = \frac{(2\sqrt{3})^6 + 0}{2^6 + 0} = \frac{64 \cdot 27}{64} = 27$.

Evaluate the limit: $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$

Similar Questions

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 2} \frac{3x^{2}-x-10}{x^{2}-4}$

$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} = $

$\lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$ is equal to

The value of $\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^{2}}$ is

$\lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module