Question

How do you find the limit of $ \left( {{x^2} + {x^3}} \right) $ as $ x $ approaches infinity?

Answer 1

Hint: In order to determine the limit of the above function, first whether the function can be more simplified or not. If the function can be simplified then do it or if it cannot be further simplified then just put the limit value in the function and get the results. As $ x $ approaches to infinity $ {x^2} $ will also approach infinity.

Complete step by step solution:
We are given a function in variable $ x $ i.e. $ \left( {{x^2} + {x^3}} \right) $ having limit $ x \to \infty $ .
 $ \mathop {\lim }\limits_{x \to \infty } \left( {{x^2} + {x^3}} \right) $
Since, we know that before proceeding ahead always check that the function can be modified to its simplest form or not.
In this function we can see that there is no need of modifying as it is already in its simplest form.
So just simply put the limit value in place of $ x $ and we get:
 $
  \mathop {\lim }\limits_{x \to \infty } \left( {{x^2} + {x^3}} \right) \\
  \mathop {\lim }\limits_{x \to \infty } \left( {{\infty ^2} + {\infty ^3}} \right) \;
  $
We know that Square or cube of \[\infty \] will give \[\infty \] only.
Hence, the above result is completely inconsistent.
Therefore, the limit of $ \left( {{x^2} + {x^3}} \right) $ as $ x \to \infty $ is equal to \[\infty \].
So, the correct answer is “\[\infty \]”.

Note: 1. Don’t forget to cross-check your answer.
2.After putting the Limit the result should never in the indeterminate form $ \dfrac{0}{0}\,or\,\dfrac{{ \pm \infty }}{{ \pm \infty }} $ . If it is contained, apply some operation to modify the result or use the L-Hospitals rule.
3. Avoid any step jump in such types of questions as this might increase the chances of mistakes.