Question

What is the limit of $\dfrac{1}{\sqrt{x}}$ as it goes to infinity?

Answer 1

Hint: The limit of a function for a point that lies in its domain is the value that the function may exhibit as the value approaches the point and this is the most fundamental concept of calculus. We know that $\dfrac{1}{\infty }=0$ and the square root of infinity is always infinity. The limit is represented as $\displaystyle \lim_{x \to \infty }$ where the values listed are changed according to the requirement.

Complete step by step answer:
Given,
$\dfrac{1}{\sqrt{x}}$
To find the limit of $\dfrac{1}{\sqrt{x}}$ as it goes to infinity.
We can write the given condition in terms of expression:
$\displaystyle \lim_{x \to \infty }\dfrac{1}{\sqrt{x}}$
The value of $x$ has to be positive since its square root is taken and we observe that we get:
$x \to \infty $ (Since this value is non-negative, increasing and not bounded)
Thus, we have
$\dfrac{1}{\sqrt{x}}$ can be made as close to zero as we want.
Thus, we choose $\varepsilon $ such that we have $\varepsilon >0$.
Since $\sqrt{x}$ is an increasing function we have that the limit of the function approaches zero as the value of $x$ approaches infinity.
Therefore, we have
$\displaystyle \lim_{x \to \infty }\dfrac{1}{\sqrt{x}}=0$

Note: There is another method to solve this question which is much simpler than discussed above which is by discussing the basic definition of limit and square root of a function since square root of a number is always a positive number therefore, we can write:
$\displaystyle \lim_{x \to \infty }\dfrac{1}{\sqrt{x}}$
As the above value of $x$ approaches infinity we have by basic definition that $\dfrac{1}{\infty }=0$ therefore the above equation becomes:
$\displaystyle \lim_{x \to \infty }\dfrac{1}{\sqrt{\infty }}\to 0$ as $x \to \infty $
Therefore, we see that the value obtained in either case is the same.