   Chapter 5.6, Problem 7SWU ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# In Exercises 7-10, find the limit. lim x → ∞ 2 x 2 + 4 x − 1 3 x 2 − 2 x

To determine

To calculate: The value of the limit limx2x2+4x13x22x.

Explanation

Given Information:

The provided limit is limx2x2+4x13x22x.

Formula used:

Properties of limits:

limxaf(x)g(x)=limxaf(x)limxag(x)

limxa(f(x)±g(x))=limxaf(x)±limxag(x)

Special limits limx1x=0 and limx1x2=0

Calculation:

Consider the provided limit,

limx2x2+4x13x22x

Dividing the numerator and denominator throughout by x2

limx2x2+4x1x23x22xx2=limx2+4x1x232x

Applying the limit to both the numerator and denominator using limxaf(x)g(x)=limxaf(x)limxag(x) and then apply limxa(f(x)±g(x)

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