# Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

The value of the algebraic function is infinity as $x$ approaches infinity. So, an alternate mathematical approach should be used to solve this limit problem.

### Use Rationalization method

Multiply and divide the function $\sqrt{x^2+x+1}-\sqrt{x^2+1}$ by its conjugate function.

$= \displaystyle \large \lim_{x \,\to\, \infty} \normalsize {(\sqrt{x^2+x+1}-\sqrt{x^2+1})}$ $\times$ $\dfrac{\sqrt{x^2+x+1}+\sqrt{x^2+1}}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}$

### Multiply conjugate functions

Use expansion of (a+b)(a-b) identity to multiply the conjugate functions mathematically.

$= \displaystyle \large \lim_{x \,\to\, \infty} \normalsize \dfrac{{(\sqrt{x^2+x+1}-\sqrt{x^2+1})}{(\sqrt{x^2+x+1}+\sqrt{x^2+1})}}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}$

$= \displaystyle \large \lim_{x \,\to\, \infty} \normalsize \dfrac{{(\sqrt{x^2+x+1})}^2-{(\sqrt{x^2+1})}^2}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}$

### Simplify the Limit of the function

Now simplify the limit of the function to the maximum level.

$= \displaystyle \large \lim_{x \,\to\, \infty} \normalsize \dfrac{(x^2+x+1)-(x^2+1)}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}$

$= \displaystyle \large \lim_{x \,\to\, \infty} \normalsize \dfrac{x^2+x+1-x^2-1}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}$

$= \displaystyle \large \lim_{x \,\to\, \infty} \normalsize \require{cancel} \dfrac{\cancel{x^2}+x+\cancel{1}-\cancel{x^2}-\cancel{1}}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}$

$= \displaystyle \large \lim_{x \,\to\, \infty} \normalsize \dfrac{x}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}$

Express the function in reciprocal form for simplifying the function further.

$= \displaystyle \large \lim_{x \,\to\, \infty} \normalsize \dfrac{1}{\dfrac{\sqrt{x^2+x+1}+\sqrt{x^2+1}}{x}}$

$= \dfrac{1}{\displaystyle \large \lim_{x \,\to\, \infty} \normalsize \dfrac{\sqrt{x^2+x+1}+\sqrt{x^2+1}}{x}}$

$= \dfrac{1}{\displaystyle \large \lim_{x \,\to\, \infty} \normalsize \Bigg[\dfrac{\sqrt{x^2+x+1}}{x}+\dfrac{\sqrt{x^2+1}}{x}\Bigg]}$

$= \dfrac{1}{\displaystyle \large \lim_{x \,\to\, \infty} \normalsize \Bigg[\dfrac{\sqrt{x^2+x+1}}{\sqrt{x^2}}+\dfrac{\sqrt{x^2+1}}{\sqrt{x^2}}\Bigg]}$

$= \dfrac{1}{\displaystyle \large \lim_{x \,\to\, \infty} \normalsize \Bigg[\sqrt{\dfrac{x^2+x+1}{x^2}}+\sqrt{\dfrac{x^2+1}{x^2}}\Bigg]}$

$= \dfrac{1}{\displaystyle \large \lim_{x \,\to\, \infty} \normalsize \Bigg[\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}\Bigg]}$

Use quotient rule of exponents for simplifying the function to the final level.

$= \dfrac{1}{\displaystyle \large \lim_{x \,\to\, \infty} \normalsize \require{cancel} \Bigg[\sqrt{\dfrac{\cancel{x^2}}{\cancel{x^2}}+\dfrac{\cancel{x}}{\cancel{x^2}}+\dfrac{1}{x^2}}+\sqrt{\dfrac{\cancel{x^2}}{\cancel{x^2}}+\dfrac{1}{x^2}}\Bigg]}$

$= \dfrac{1}{\displaystyle \large \lim_{x \,\to\, \infty} \normalsize \Bigg[\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}+\sqrt{1+\dfrac{1}{x^2}}\Bigg]}$

### Evaluate the function

Find the value of the function as $x$ approaches infinity.

$= \dfrac{1}{\sqrt{1+\dfrac{1}{\infty}+\dfrac{1}{{(\infty)}^2}}+\sqrt{1+\dfrac{1}{{(\infty)}^2}}}$

$= \dfrac{1}{\sqrt{1+0+0}+\sqrt{1+0}}$

$= \dfrac{1}{1+1}$

$= \dfrac{1}{2}$

Email subscription
Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects. Know more