limx→∞ x2+x+1−x2+1= - Infinity Learn by Sri Chaitanya

A
$- \frac{1}{2}$
B
$\frac{1}{2}$
C
1
D
-1

Solution:

Here, the expression assumes the form $\infty - \infty$ as

$x \to \infty$ .So, we reduce it to the rational form $\frac{f (x)}{g (x)}$

We have,

$\begin{array}{l} lim_{x \to \infty} \sqrt{x^{2} + x + 1} - \sqrt{x^{2} + 1} \\ = lim_{x \to \infty} \frac{\{\sqrt{x^{2} + x + 1} - \sqrt{x^{2} + 1}\} \{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}\}}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}} \\ = lim_{x \to \infty} \frac{x^{2} + x + 1 - x^{2} - 1}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}} = lim_{x \to \infty} \frac{x}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}} \end{array}$

$= lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^{2}}} + \sqrt{1 + \frac{1}{x^{2}}}}$ [Dividing $N$ , and D' by $x$ ]

$= \frac{1}{1 + 1} = \frac{1}{2}$

$lim_{x \to \infty} (\sqrt{x^{2} + x + 1} - \sqrt{x^{2} + 1}) =$

Solution:

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