Determine the infinite limit. Need Help? Read It Talk to a Tutor 4. O points SCa

Question

Determine the infinite limit. Need Help? Read It Talk to a Tutor 4. O points SCalcET8 23 010 (a) what is wrong with the following equation? (x 2)(x 3) x2 x 6 The left-hand side is not defined for x 0, but the right-hand side is The left-hand side is not defined for x 2, but the right-hand side is None of these the equation is correct. (b) In view of part (a), explain why the equation lim (x 3) is correct. and x 3 are both continuous, the equation follows. Since since the equation holds for all x 2, it follows that both sides of the equation approach the same limit as x 2. This equation follows from the fact that the equation in part (a) is correct. None of these the equation is not correct

Explanation / Answer

we have given lim x-->3- (e^x/(x-3)^3)

lim x-->3- (e^x/(x-3)^3) =[lim x-->3- (e^x)] [lim x-->3-(1/(x-3)^3)]

=e^3* [lim x-->3-(1/(x-3)^3)]

[lim x-->3-(1/(x-3)^3)]

for x approaching 3 ,x<3 implies (x-3)^3<0

the denominator is negative quantity approaching 0 from the left

lim x-->3-(1/(x-3)^3)=-infinity

lim x-->3- (e^x/(x-3)^3) =e^3* [lim x-->3-(1/(x-3)^3)]=e^3*(-infinity)=-infinity

lim x-->3- (e^x/(x-3)^3) =-infinity

a) we have given (x^2+x-6)/(x-2) =x+3

x^2+x-6=(x-2)(x+3)

The equation is correct

b) lim x-->2 (x^2+x-6)/(x-2) = lim x-->2 (x-2)(x+3)/(x-2) since we taking out factors of (x^2+x-6)

= lim x-->2 (x+3) since get cancelled (x-2) in numerator and denominator

so it is correct

we have given lim x-->a [f(x)+g(x)]=4 --(1) and lim x-->a [f(x)-g(x)]=1 ---(2)

lim x-->a f(x)+ lim x-->ag(x)=4

lim x-->a f(x)- lim x-->ag(x)=1

adding the limits of equation 1,2

2*[lim x-->a f(x)]=5 implies lim x-->a f(x)=5/2

subtracting the limits of equation 1-2

2*[lim x-->ag(x)]=3 implies lim x-->ag(x)=3/2

so lim x-->a [f(x)g(x)]= [lim x-->a f(x))] [lim x-->a g(x)] =(5/2)*(3/2) =15/4

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