Math Doubts

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.

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

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

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

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

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

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

$= \,\,\, 1$



Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more