Question

Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\] 

Answer 1

\[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\text{ LHS }: \]
\[ \lim_{x \to \infty} \left( \left( \sqrt{x^2 + x + 1} - x \right) \right)\]
\[\text{ Rationalising the numerator }: \]
\[ \lim_{x \to \infty} \left[ \frac{\left( \sqrt{x^2 + x + 1} - x \right) \left( \sqrt{x^2 + x + 1} + x \right)}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\left( x^2 + x + 1 \right) - x^2}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{x + 1}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]\]
\[\text{ Dividing the numerator and the denominator by x }: \]
\[ \lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\frac{\sqrt{x^2 + x + 1}}{x} + 1} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\sqrt{\frac{x^2 + x + 1}{x^2}} + 1} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + 1} \right]\]
\[\text{ When x } \to \infty , \text{ then } \frac{1}{x} \to 0 . \]
\[\frac{1}{\sqrt{1} + 1}\]
\[ = \frac{1}{2}\]
\[RHS: \]
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right) \left[ \text{ from }  \infty - \infty \right]\]

Rationalising the numerator: 

\[\lim_{x \to \infty} \left[ \frac{\left( \sqrt{x^2 + 1} - x \right) \left( \sqrt{x^2 + 1} + x \right)}{\left( \sqrt{x^2 + 1} + x \right)} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{x^2 + 1 - x^2}{\left( \sqrt{x^2 + 1} + x \right)} \right]\]
\[ = \frac{1}{\infty}\]
\[ = 0\]
\[ \therefore \lim_{x \to \infty} \left[ \sqrt{x^2 + x + 1} - x \right] \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]