Question

What is the limit as x approaches 0 from the right of $\dfrac{1}{x}$ ?

Answer 1

Hint: We need to calculate the right hand limit of $\dfrac{1}{x}$ as $x$ approaches 0, i.e., $\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{x}$. We must substitute small values of $x$ and try to figure out the limiting value. We will see that as $x$ approaches 0 from the positive side, $\dfrac{1}{x}$ approaches positive infinity.

Complete step-by-step solution:
We know that by calculating limits at a point, we try to calculate the value of a function that may or may not be defined at a particular value. $\displaystyle \lim_{x \to 0}f\left( x \right)$ is the value of $f\left( x \right)$ as the value of $x$ gets very close to 0.
We know that limits are of two types, left hand limit and right hand limit, which are expressed as $\displaystyle \lim_{x \to {{0}^{-}}}f\left( x \right)$ and $\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right)$ respectively.
We know that in left hand limit, the value of $x$ gets closer and closer to 0 from the left side on the number line, i.e., the value of $x$ is always less than 0.
We are also aware that in right hand limit, the value of $x$ gets closer and closer to 0 from the right side on the number line, i.e., the value of $x$ is always greater than 0.
To calculate the limit as $x$ approaches 0 from the right side of $\dfrac{1}{x}$ means that we need to calculate $\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{x}$ .
Let us put some positive values of $x$ closer to 0 and check what happens.
If $x=1$, the value of $\dfrac{1}{x}$ will be equal to 1.
If $x=\dfrac{1}{2}$, the value of $\dfrac{1}{x}$ will be equal to 2.
If $x=\dfrac{1}{5}$, the value of $\dfrac{1}{x}$ will be equal to 5.
If $x=\dfrac{1}{10}$, the value of $\dfrac{1}{x}$ will be equal to 10.
If $x=\dfrac{1}{100}$, the value of $\dfrac{1}{x}$ will be equal to 100.
If $x=\dfrac{1}{1000}$, the value of $\dfrac{1}{x}$ will be equal to 1000.
Here, we can clearly see that as the values are getting closer and closer to 0, the value of $\dfrac{1}{x}$ is getting larger and larger.
So, when $x={{0}^{+}}$, i.e., $x$ is very close to zero, but slightly greater than 0, then $\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{x}=\dfrac{1}{h}$ where h is an infinitesimal small positive number. And so, $\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{x}=+\infty $.
Hence, the limit as $x$ approaches 0 from the right of $\dfrac{1}{x}$ is positive infinity.

Note: We must be careful not to use L’ Hospital’s rule in this problem. We must remember that L’ Hospital’s rule is applicable only in the case of indeterminate forms, and $\dfrac{1}{0}$ is not an indeterminate form.