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If lim(f+g) = 1 as x -> a and lim(f-g) = 1 as x -> a. What is lim fg as x -> a ?

Question:

Given lim{eq}_{x \to a } {/eq} (f+g) = 1 and lim{eq}_{x \to a } {/eq} (f-g) = 1.

What is lim{eq}_{x \to a} {/eq} fg ?

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}.

if {eq}\mathop {\lim }\limits_{x \to {t^ + }} a(x) = {A_1} {/eq} and {eq}\mathop {\lim }\limits_{x \to {t^ - }} a(x) = {A_2} {/eq}, then;

(1) Limit exist if,

{eq}{A_1} = {A_2} {/eq}

(2) Limit does not exist if,

{eq}{A_1} \ne {A_2} {/eq}

Answer and Explanation:

{eq}\displaystyle \eqalign{ & \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = 1 \cr & \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - g\left( x \right)} \right) = 1 \cr & \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = 1\,\,\,\,\, \Rightarrow \left( {\mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr & \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - g\left( x \right)} \right) = 1\,\,\,\,\, \Rightarrow \left( {\mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right) \cr & \cr & {\text{Adding }}\left( 1 \right){\text{ and }}\left( 2 \right); \cr & \left( {\mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right) + \left( {\mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right) = 1 + 1 \cr & 2\mathop {\lim }\limits_{x \to a} f\left( x \right) = 2\,\,\,\, \Rightarrow \mathop {\lim }\limits_{x \to a} f\left( x \right) = 1 \cr & \cr & {\text{From (1);}} \cr & \left( {\mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right) = 1\,\,\, \Rightarrow \left( {1 + \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right) = 1 \cr & \mathop {\lim }\limits_{x \to a} g\left( x \right) = 1 - 1 = 0\,\,\, \Rightarrow \mathop {\lim }\limits_{x \to a} g\left( x \right) = 0 \cr & \cr & \mathop {\lim }\limits_{x \to a} f\left( x \right)g\left( x \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right)\mathop {\lim }\limits_{x \to a} g\left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From limit properties}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( 1 \right)\left( 0 \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0 \cr & \cr & \mathop {\lim }\limits_{x \to a} f\left( x \right)g\left( x \right) = 0 \cr} {/eq}


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Understanding the Properties of Limits

from Math 104: Calculus

Chapter 6 / Lesson 5
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