Saturday, October 12, 2019

Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 2

Evaluate each limit if it exists. If the limit does not exist, explain why. Use the given graphs of f and g





(a) limx2[f(x)+g(x)](b) limx1[f(x)+g(x)](c) limx0[f(x)g(x)](d) limx1f(x)g(x)(e) limx2[x3f(x)](f) limx23+f(x)


a.) limx2[f(x)+g(x)]

limx2[f(x)+g(x)]=limx2f(x)+limx2g(x)(Substitute the values of f(x) and g(x))limx2[f(x)+g(x)]=2+0(Simplify)limx2[f(x)+g(x)]=2


b.) limx1[f(x)+g(x)]

limx1[f(x)+g(x)]=limx1f(x)+limx1g(x)(Substitute the values of f(x) and g(x))limx1f(x)=1limx1g(x)=Does not existThe given equation does not exist because the left and the right limits of the function g(x) are different.


c.) limx0[f(x)g(x)]

limx0[f(x)g(x)]=limx0f(x)limx0g(x)(Substitute the values of f(x) and g(x))limx0[f(x)g(x)]=(0)(1.5)(Simplify)limx0[f(x)g(x)]=0


d.) limx1f(x)g(x)

limx1f(x)g(x)=limx1f(x)limx1g(x)(Substitute the values of f(x) and g(x))limx1f(x)g(x)=10(Does not exist)The limit does not exist, the function is undefined because the denominator is zero.



e.) limx2[x3f(x)]

limx2[x3f(x)]=limx2x3limx2f(x)limx2[x3f(x)]=(2)3(2)limx2[x3f(x)]=16


f.) limx23+f(x)

limx23+f(x)(Substitute the value of f(x))limx23+f(x)=3+2(Simplify)limx23+f(x)=5

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