Question

What is the limit of $\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}$ as $x\to \infty$?

Answer 1

We express the limit in a different way where we take the help of the limit value $x\to \infty$ in the form of $\dfrac{1}{x}\to 0$. We divide both the numerator and denominator with ${{x}^{2}}$. We put the limit value and find the solution of the expression.

Complete step by step answer:
We have to find the limit of $\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}$. We simplify the equation to place the limit value by dividing both the numerator and denominator with ${{x}^{2}}$.

\Rightarrow \dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=\dfrac{x-\dfrac{2}{x}+\dfrac{3}{{{x}^{2}}}}{\dfrac{5}{{{x}^{2}}}-2}$$ Now the limit value is given as $x\to \infty $. Therefore, $\dfrac{1}{x}\to 0$. We can form the expression as $$\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=\dfrac{x-2\left( \dfrac{1}{x} \right)+3{{\left( \dfrac{1}{x} \right)}^{2}}}{5{{\left( \dfrac{1}{x} \right)}^{2}}-2}$$ We put the limit value to get $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=\underset{\dfrac{1}{x}\to 0}{\mathop{\lim }}\,\dfrac{x-2\left( \dfrac{1}{x} \right)+3{{\left( \dfrac{1}{x} \right)}^{2}}}{5{{\left( \dfrac{1}{x} \right)}^{2}}-2} \\\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=-\dfrac{x}{2} \\\ \therefore \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=-\infty $ **Therefore, the limit of $\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}$ as $x\to \infty $ is $-\infty $.** **Note:** We can also use the L’Hospital rule where we take the differentiated form of the numerator and denominator until it gives any other form of limit other than $\dfrac{\infty }{\infty },\dfrac{0}{0}$. Here we get $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=\dfrac{\infty }{\infty }$. So, we differentiate. $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{3{{x}^{2}}-2}{-4x}$. It is still in $\dfrac{\infty }{\infty }$ form. So, $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{3{{x}^{2}}-2}{-4x}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{6x}{-4}$. Now we can put the limit value directly to get $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}-2x+3}{5-2{{x}^{2}}}=-\infty $.