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Question: What is the limit as x approaches 0 from the right of \(\dfrac{1}{x}\) ?...

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

Explanation

Solution

We need to calculate the right hand limit of 1x\dfrac{1}{x} as xx approaches 0, i.e., limx0+1x\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{x}. We must substitute small values of xx and try to figure out the limiting value. We will see that as xx approaches 0 from the positive side, 1x\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. limx0f(x)\displaystyle \lim_{x \to 0}f\left( x \right) is the value of f(x)f\left( x \right) as the value of xx gets very close to 0.
We know that limits are of two types, left hand limit and right hand limit, which are expressed as limx0f(x)\displaystyle \lim_{x \to {{0}^{-}}}f\left( x \right) and limx0+f(x)\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right) respectively.
We know that in left hand limit, the value of xx gets closer and closer to 0 from the left side on the number line, i.e., the value of xx is always less than 0.
We are also aware that in right hand limit, the value of xx gets closer and closer to 0 from the right side on the number line, i.e., the value of xx is always greater than 0.
To calculate the limit as xx approaches 0 from the right side of 1x\dfrac{1}{x} means that we need to calculate limx0+1x\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{x} .
Let us put some positive values of xx closer to 0 and check what happens.
If x=1x=1, the value of 1x\dfrac{1}{x} will be equal to 1.
If x=12x=\dfrac{1}{2}, the value of 1x\dfrac{1}{x} will be equal to 2.
If x=15x=\dfrac{1}{5}, the value of 1x\dfrac{1}{x} will be equal to 5.
If x=110x=\dfrac{1}{10}, the value of 1x\dfrac{1}{x} will be equal to 10.
If x=1100x=\dfrac{1}{100}, the value of 1x\dfrac{1}{x} will be equal to 100.
If x=11000x=\dfrac{1}{1000}, the value of 1x\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 1x\dfrac{1}{x} is getting larger and larger.
So, when x=0+x={{0}^{+}}, i.e., xx is very close to zero, but slightly greater than 0, then limx0+1x=1h\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{x}=\dfrac{1}{h} where h is an infinitesimal small positive number. And so, limx0+1x=+\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{x}=+\infty .
Hence, the limit as xx approaches 0 from the right of 1x\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 10\dfrac{1}{0} is not an indeterminate form.