The true statement for mathop {lim }limits_{x | vedclass

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

Question

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$.Rationalizing the numerator

Vedclass · Accepted Answer

Lies between $0$ and $\frac{1}{2}$

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$.Rationalizing the numerator and denominator:$L = \mathop {\lim }\limits_{x 	o 0} \frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})}{(\sqrt{2+3x} - \sqrt{2-3x})(\sqrt{2+3x} + \sqrt{2-3x})} 	imes \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{(1+x) - (1-x)}{(2+3x) - (2-3x)} 	imes \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{2x}{6x} 	imes \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$$L = \frac{1}{3} 	imes \frac{\sqrt{2} + \sqrt{2}}{1 + 1} = \frac{1}{3} 	imes \frac{2\sqrt{2}}{2} = \frac{\sqrt{2}}{3}$.Since $\sqrt{2} \approx 1.414$,then $\frac{\sqrt{2}}{3} \approx \frac{1.414}{3} \approx 0.471$.Thus,$0 Therefore,the correct option is $B$.

The true statement for $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$ is

Similar Questions

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

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^3} + 1}} + \frac{4}{{{n^3} + 1}} + \frac{9}{{{n^3} + 1}} + \dots + \frac{{{n^2}}}{{{n^3} + 1}}} \right] = $

Let $p = \mathop {\lim }\limits_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$,then $\log p = $ . . .

If $0 < p < q$,then $\lim _{n \rightarrow \infty}\left(q^n+p^n\right)^{1 / n}$ is equal to :

If $S_1 = \sum_{r=1}^{n} r$,$S_2 = \sum_{r=1}^{n} r^2$,and $S_3 = \sum_{r=1}^{n} r^3$,then the value of $\lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module