# 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.

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

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

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

$= \dfrac{1}{2}$

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