# Let f(x) \text{ and } g(x) be functions defined on a set A and assume f(x) \leq g(x)...

## Question:

Let {eq}f(x) \text{ and } g(x) {/eq} be functions defined on a set {eq}A {/eq} and assume {eq}f(x) \leq g(x) \ \forall x\in A {/eq}. Furthermore, let {eq}c {/eq} be a limit point of {eq}A {/eq} and assume that limits for {eq}f(x) \text{ and } g(x) {/eq} exist at {eq}c {/eq}. Show that

{eq}\lim\limits_{x\to c} f(x) \leq \lim\limits_{x\to c} g(x) {/eq}.

## Limits:

To determine the limit for a function, the behavior of the expression that defines the function must be analyzed when the input value of choice is approached.

## Answer and Explanation:

Given: {eq}\lim_{x\to c} f(x) \leq \lim_{x\to c} g(x) {/eq}

Given the statement that {eq}f(x) \leq g(x) \forall x\in A {/eq}, where {eq}\forall x\in A {/eq} means specifically that all {eq}x {/eq} values in the set {eq}A {/eq} satisfy the condition that {eq}f(x) \leq g(x) {/eq}. Since it is mentioned that the limits for {eq}f(x) \text{ and } g(x) {/eq} exist at {eq}c {/eq} and the expressions that define each function are not given it can be assumed that both functions are continuous at the input value {eq}c {/eq}. Therefore, direct substitution of the input value can be utilized for each limit.

{eq}\begin{align*} \lim_{x\to c} f(x) \leq \lim_{x\to c} g(x) &\Rightarrow f(c) \leq g(c) \\ \end{align*} {/eq}

Since the input value {eq}c {/eq} is a limit point of {eq}A {/eq} and all input values in the set {eq}A {/eq} follow the condition {eq}f(x) \leq g(x) {/eq}, it has been proved that {eq}\lim_{x\to c} f(x) \leq \lim_{x\to c} g(x) {/eq}.

#### Learn more about this topic:

How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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