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

Let us try to evaluate the limit of the algebraic function $\sqrt{x^2+2x}-x$ as $x$ approaches infinity by the direct substitution method.

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

$= \,\,\,$ $\sqrt{{(\infty)}^2+2(\infty)}-(\infty)$

$= \,\,\, \infty$

The limit of algebraic function as $x$ approaches infinity is undefined. So, try to find the limit of the function in another method.

### Rationalize the function

The algebraic function is in radical form and the limit of this function is undefined as $x$ approaches infinity. So, the limit of this radical function can be calculated by using rationalization method by multiplying and dividing the function by its conjugate function.

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

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

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

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

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

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

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

### Simplify the function

$x$ is a factor in the numerator and take $x$ common from all the terms of the denominator for making them to get cancelled.

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

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

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

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

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

### Evaluate Limit of the function

Now, evaluate the limit of the algebraic function as $x$ approaches infinity.

$= \,\,\,$ $\dfrac{2}{\sqrt{1+\dfrac{2}{\infty}}+1}$

$= \,\,\,$ $\dfrac{2}{\sqrt{1+0}+1}$

$= \,\,\,$ $\dfrac{2}{\sqrt{1}+1}$

$= \,\,\,$ $\dfrac{2}{1+1}$

$= \,\,\,$ $\dfrac{2}{2}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{2}}{\cancel{2}}$

$= \,\,\, 1$

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