Question

Compute $\displaystyle \lim_{x \to \infty }\sqrt{{{x}^{2}}+x}-x$.

Answer 1

Hint: To solve this question we need to have the knowledge of the limit, which will help us in determining the limit of the given function. Now the first step will be to check the determinant form of the function when the limit of “x” is substituted in the function. Now, on the basis of the limit the function is solved accordingly.

Complete step-by-step solution:
The question ask us to find the limit of the function given when X approaches Infinity which is mathematically written as $\displaystyle \lim_{x \to \infty }\sqrt{{{x}^{2}}+x}-x$. We will first find the indeterminate form of the function by substituting the value of “x” in the above function. On doing so we get:
$\Rightarrow \sqrt{\infty +\infty }-\infty $
On writing it more clearly we get:
$\Rightarrow \infty -\infty $
This means the initial form of the limit is $\infty -\infty $.
The second step is to solve for the function. For this we will use conjugate. The same function will be multiplied to both the numerator and the denominator. So the function on taking the conjugate we get:
$\Rightarrow \sqrt{{{x}^{2}}+x}-x=\dfrac{\sqrt{{{x}^{2}}+x}-x}{1}\times \dfrac{\left( \sqrt{{{x}^{2}}+x}+x \right)}{\sqrt{{{x}^{2}}+x}+x}$
On solving the above expression we get:
$\Rightarrow \dfrac{\sqrt{{{x}^{2}}+x}-x}{1}\times \dfrac{\left( \sqrt{{{x}^{2}}+x}+x \right)}{\sqrt{{{x}^{2}}+x}+x}$
$\Rightarrow \dfrac{\left( {{\left( \sqrt{{{x}^{2}}+x} \right)}^{2}}-{{x}^{2}} \right)}{\left( \sqrt{{{x}^{2}}+x}+x \right)}$
$\Rightarrow \dfrac{\left( {{x}^{2}}+x-{{x}^{2}} \right)}{\left( \sqrt{{{x}^{2}}+x}+x \right)}$
On simplifying the numerator we get:
$\Rightarrow \dfrac{x}{\left( \sqrt{{{x}^{2}}+x}+x \right)}$
We will again check the form of the function on applying the limit. So doing this we get:
 \[\Rightarrow \dfrac{\infty }{\infty }\]
This indeterminate form can be factored and reduced. Now, we are aware of the fact that $\sqrt{{{x}^{2}}}=\left| x \right|$, so for the positive value of $x$ the function will behave as:
$\Rightarrow \dfrac{x}{\left( \sqrt{{{x}^{2}}+x}+x \right)}=\dfrac{x}{\sqrt{{{x}^{2}}}\sqrt{1+\dfrac{1}{x}}+1}$
The function is applicable for all the values of $x$ except $0$. On solving further we get:
$\Rightarrow \dfrac{x}{x\sqrt{1+\dfrac{1}{x}}+1}$
$\Rightarrow \dfrac{1}{\sqrt{1+\dfrac{1}{x}}+1}$
Now we will put the limit in the above expression. On doing so we get:
$\Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{1}{\sqrt{1+\dfrac{1}{x}}+1}$
$\Rightarrow \dfrac{1}{\sqrt{1+\dfrac{1}{\infty }}+1}$
Since we know that $\dfrac{1}{\infty }=0$ , on putting this in the equation we get:
$\Rightarrow \dfrac{1}{\sqrt{1+0}+1}$
On further calculation we get:
$\Rightarrow \dfrac{1}{2}$
$\therefore $ The value of $\displaystyle \lim_{x \to \infty }\sqrt{{{x}^{2}}+x}-x$ is $\dfrac{1}{2}$ .

Note:To solve this type of question the first step should always be to check the indeterminate form. In the number system in total we have $7$ indeterminate forms. Conjugate of a function is formed by changing the sign between the terms in a binomial function. Example: Conjugate of $x+y$ is $x-y$.