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The value of limit \lim _{x \rightarrow \infty}(x-\sqrt{x^{2}+x}) is

Option: 1 -0.5

Option: 2 0.5

Option: 3 1

Option: 4 -1

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Limit of the type  ∞ - ∞ -

Limit of the type  ∞ - ∞

This type of problems can be solved by rationalization

Let’s go through the illustration to understand how to solve such type of questions

 

The value of the limit \lim _{x \rightarrow \infty}(\sqrt{(x+2 a)(2 x+a)}-x \sqrt{2}) is

 

First Rationalising the expression

 

\\\Rightarrow \lim _{x \rightarrow \infty}\frac{\left (\sqrt{(x+2 a)(2 x+a)}-x \sqrt{2} \right )\left (\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2} \right )}{\left (\sqrt{(x+2 a)(2 x+a)}-x \sqrt{2} \right )}\\\\\Rightarrow \lim _{x \rightarrow \infty} \frac{(x+2 a)(2 x+a)-2 x^{2}}{(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})}\\\\\Rightarrow \lim _{x \rightarrow \infty} \frac{2x^2+5ax+2a^2-2 x^{2}}{(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})}\\\\\Rightarrow \lim _{x \rightarrow \infty} \frac{5 a x+2 a^{2}}{\sqrt{2 x^{2}+5 a x+2 a^{2}}+x \sqrt{2}}\\\\\text{Dividing numerator and denominator by x, we get :}\\\\\Rightarrow \lim _{x \rightarrow \infty} \frac{5 a+\frac{2 a^{2}}{x}}{\sqrt{2+\frac{5 a}{x}+\frac{2 a^{2}}{x^{2}}}+\sqrt{2}}=\frac{5 a}{2 \sqrt{2}}




 

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\begin{array}{l}{\lim _{x \rightarrow \infty}(x-\sqrt{x^{2}+x})} \\ {\text { On rationalising the expression, we get: }} \\ {=\lim _{x \rightarrow \infty} \frac{x^{2}-\left(x^{2}+x\right)}{x+\sqrt{x^{2}+x}}=\lim _{x \rightarrow \infty} \frac{-x}{x+\sqrt{x^{2}+x}}}\end{array}

\begin{array}{l}{\text { Divide by the highest power of } x \text { i.e. } x} \\ {\qquad \begin{aligned}=& \lim _{x \rightarrow \infty} \frac{-1}{1+\sqrt{1+\frac{1}{x}}}=\frac{-1}{1+1}=-\frac{1}{2} \end{aligned}}\end{array}

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