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Chapter 14: Q. 42 (page 883)

In Problems 7鈥� 42, find each limit algebraically.

limx鈫�3x3-3x2+4x-12x4-3x3+x-3.

Short Answer

Expert verified

The answer is1328.

Step by step solution

01

Step. 1 Given Information

Firstly, we check whether the given function is in indeterminant form or not.

f(x)=x3-3x2+4x-12x4-3x3+x-3

Put x=3in the numerator we get,

x3-3x2+4x-12=27-3(9)+4(3)-12=27-27+12-12=0.

Put x=3in the denominator we get,

x4-3x3+x-3=81-3(27)+3-3=81-81+3-3=0.

Since both numerator and denominator gives 0 means they both have x=3as their common root.

02

Step. 2 Factorizing

Numerator , x3-3x2+4x-12=x2(x-3)+4(x-3)=(x2+4)(x-3).

Denominator,x4-3x3+x-3=x3(x-3)+1(x-3)=(x3+1)(x-3).

So,

limx鈫�3x3-3x2+4x-12x4-3x3+x-3=limx鈫�3(x2+4)(x-3)(x3+1)(x-3)=limx鈫�3x2+4x3+1.

Now we can put the limit directly.

03

Step. 3 Final calculation of the limit

limx鈫�3x2+4x3+1=9+427+1=1328.

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