# Which of the following functions grows the fastest as x goes to infinity? (a) f (x) = e^x. (b) f...

## Question:

Which of the following functions grows the fastest as {eq}x \to \infty {/eq}?

(a) {eq}f (x) = e^x {/eq}.

(b) {eq}f (x) = 2 x {/eq}.

(c) {eq}f (x) = 4 x {/eq}.

(d) {eq}\displaystyle f(x) = e^{-2 x} {/eq}.

## Existence of a Limit:

Suppose that {eq}a(x) {/eq} is a function that is defined in an interval that contains {eq}x=t {/eq}. Then limit is defined as:

{eq}\mathop {\lim }\limits_{x \to t} a(x) = M {/eq}

There exists a very small number {eq}k {/eq} such that {eq}k>0 {/eq} so that {eq}n>0 {/eq}. This means that {eq}|a(x) - M| < k {/eq} whenever {eq}0 < |x - t| < n {/eq}.

## Answer and Explanation:

{eq}\displaystyle \eqalign{ & (a) \cr & f(x) = {e^x} \cr & \mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } {e^x} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \infty \cr & \cr & (b) \cr & f(x) = 2x \cr & \mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } 2x \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( \infty \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( \infty \right) \cr & \cr & (c) \cr & f(x) = 4x \cr & \mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } 4x \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\left( \infty \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( \infty \right) \cr & \cr & (d) \cr & f(x) = {e^{ - 2x}} \cr & \mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } {e^{ - 2x}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0 \cr & \cr & {e^x} > > 4x > 2x > > {e^{ - 2x}} \cr & \cr & {\text{Option (a) }}f(x) = {e^x}{\text{ functions grows the fastest as }}x \to \infty . \cr} {/eq}

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