mathop {lim }limits_{n to infty } frac{{sqrt | vedclass

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Question

(B) To evaluate the limit $\mathop {\lim }\limits_{n 	o \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }}$,divide both the numerator and the den

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) To evaluate the limit $\mathop {\lim }\limits_{n 	o \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }}$,divide both the numerator and the denominator by $\sqrt{n}$:$\mathop {\lim }\limits_{n 	o \infty } \frac{\frac{\sqrt{n}}{\sqrt{n}}}{\frac{\sqrt{n}}{\sqrt{n}} + \frac{\sqrt{n+1}}{\sqrt{n}}} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{1 + \sqrt{1 + \frac{1}{n}}}$As $n 	o \infty$,the term $\frac{1}{n} 	o 0$.Therefore,the limit becomes $\frac{1}{1 + \sqrt{1 + 0}} = \frac{1}{1 + 1} = \frac{1}{2}$.

$\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }} = $

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