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Question: How do you find the limit for \[{e^{ - x}}\] when \(x\) approaches infinity?...

How do you find the limit for e−x{e^{ - x}} when xx approaches infinity?

Explanation

Solution

In the given problem we have to find the 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 an expression to which limit can be applied easily.

Complete step by step answer:
Before dwelling on the question, we need to understand the behaviour of limits approaching infinity. In other words, we are going to understand what happens with the value of xx 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 xx starts increasing, the value 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 xx. 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 at hand.We have,
lim⁡x→∞e−x\mathop {\lim }\limits_{x \to \infty } {e^{ - x}}
We can rewrite this,
⇒lim⁡x→a1ex−−−−−(2)\Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{1}{{{e^x}}} - - - - - \left( 2 \right)
As k−a=1ka{k^{ - a}} = \dfrac{1}{{{k^a}}}.
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→∞1ex=0\therefore\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{e^x}}} = 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. Therefore, when any constant is multiplied with the function whose limit approaches infinity we can conclude that lim⁡x→−∞f(ax)=∞\mathop {\lim }\limits_{x \to - \infty } f\left( {ax} \right) = \infty .