Question

How do you find the limit for \[{e^{ - x}}\] when $x$ approaches infinity?

Answer 1

Hint: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 $x$ becomes very large in positive or negative sense i.e.,\[\mathop {\lim }\limits_{x \to \infty } f\left( x \right)\] and \[\mathop {\lim }\limits_{x \to - \infty } f\left( x \right)\].

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

Now let us consider the problem at hand.We have,
\[\mathop {\lim }\limits_{x \to \infty } {e^{ - x}}\]
We can rewrite this,
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{1}{{{e^x}}} - - - - - \left( 2 \right)\]
As \[{k^{ - a}} = \dfrac{1}{{{k^a}}}\].
Now, as we have concluded in (1) we know that\[\mathop {\lim }\limits_{x \to - \infty } f\left( {\dfrac{1}{x}} \right) = 0\]
Therefore, we can apply that in the above expression.
\[\therefore\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{e^x}}} = 0\], which is our required limit.

Hence, zero is the answer.

Note:Talking about the conclusion that we have made in the solution i.e., \[\mathop {\lim }\limits_{x \to - \infty } f\left( {\dfrac{1}{x}} \right) = 0\]. We are not talking about when \[x = \infty \] 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 \[\mathop {\lim }\limits_{x \to - \infty } f\left( {ax} \right) = \infty \].