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Question: Find out the limit of \[\dfrac{n}{{3n - 1}}\] as \(x\) approaches to infinity....

Find out the limit of n3n−1\dfrac{n}{{3n - 1}} as xx approaches to infinity.

Explanation

Solution

In the given problem we have to find limit for the given expression as x approaches infinity. In order to find out the limit we will, at first multiply the given function with such an algebraic expression that the given function gets transformed into such an expression to which limit can be applied easily.

Complete step by step answer:
Before dwelling into the question, it is important for us to understand the behaviour of limits approaching infinity. In other words we are going to understand what happens with the value of x becomes very large in positive or negative sense i.e.,lim⁡x→∞f(x)\mathop {\lim }\limits_{x \to \infty } f\left( x \right)and lim⁡x→−∞f(x)\mathop {\lim }\limits_{x \to - \infty } f\left( x \right).

Suppose we have,lim⁡x→−∞f(1x)\mathop {\lim }\limits_{x \to - \infty } f\left( {\dfrac{1}{x}} \right)
Now, analytically speaking, if the value of x starts increasing, the value of 1x\dfrac{1}{x} will automatically start reducing i.e., the value of 1x\dfrac{1}{x} will come nearer to zero for every increase in the value of x. Therefore, we can conclude from the above statementlim⁡x→−∞f(1x)=0−−−−−(1)\mathop {\lim }\limits_{x \to - \infty } f\left( {\dfrac{1}{x}} \right) = 0 - - - - - \left( 1 \right)

Now let us consider the problem in hand. We have,
⇒lim⁡n→∞n3n−1\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{3n - 1}}
Multiplying the numerator and denominator by numerator,
Therefore,
⇒lim⁡n→∞nn3n−1n\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{n}{n}}}{{\dfrac{{3n - 1}}{n}}}
Simplifying we get,
⇒lim⁡n→∞13−1n\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{3 - \dfrac{1}{n}}}
Now, as we have concluded in (1) we know thatlim⁡x→−∞f(1x)=0\mathop {\lim }\limits_{x \to - \infty } f\left( {\dfrac{1}{x}} \right) = 0
Therefore, we can apply that in the above expression.
∴lim⁡n→∞13−1n=0\therefore\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{3 - \dfrac{1}{n}}} = 0, which is our required limit.

Hence, zero is the answer.

Note: Talking about the conclusion that we have made in the solution i.e., lim⁡x→−∞f(1x)=0\mathop {\lim }\limits_{x \to - \infty } f\left( {\dfrac{1}{x}} \right) = 0. We are not talking about when x=∞x = \infty but we know that as the value of x increases the answer gets closer and closer to zero. Likewise, we also say that lim⁡x→−∞f(ax)=∞\mathop {\lim }\limits_{x \to - \infty } f\left( {ax} \right) = \infty .