Unit 1 Progress Check: MCQ Part B

If f is the function defined by f(x)=x−11−1x, then limx→1f(x) is equivalent to which of the following? A limx→1xlimx→1x B limx→12xlimx→12x C limx→1(x−11−x)limx→1(x−11−x) D limx→1(x−1)limx→1(1−1x)

Answer A Correct. f(x)=x−11−1x=x−1xx−1x=x−1(x−1)x=xf(x)=x−11−1x=x−1xx−1x=x−1(x−1)x=x for xx close to, but not equal to, 1. Therefore, limx→1f(x)=limx→1xlimx→1f(x)=limx→1x.

The function g is given by g(x)=1x2−4x+5. The function h is given by h(x)=2x2−8x+10x2−4x+6. If f is a function that satisfies g(x)≤f(x)≤h(x) for 0<x<5, what is limx→2f(x) ? A 0 B 1 C 2 D The limit cannot be determined from the information given.

Answer B Correct. Since limx→2g (x)=122−(4)(2)+5=11=1limx→2g (x)=122−(4)(2)+5=11=1 and limx→2h (x)=2(22)−(8)(2)+1022−(4)(2)+6=22=1limx→2h (x)=2(22)−(8)(2)+1022−(4)(2)+6=22=1, it can be concluded from the squeeze theorem that limx→2f(x)=1limx→2f(x)=1 because g(x)≤f(x)≤h(x)g(x)≤f(x)≤h(x) for 0<x<50<x<5.

Let f and g be functions such that limx→4g(x)=7 and limx→4f(x)g(x)=π. What is limx→4f(x)? A π7π7 B 7+π7+π C 7π7π D The limit cannot be determined from the information given.

Answer C Correct. Because limx→4g(x)≠0limx→4g(x)≠0, the quotient property for limits can be used, as follows: π=limx→4f(x)g(x)=limx→4f(x)limx→4g(x)=limx→4f(x)7π=limx→4f(x)g(x)=limx→4f(x)limx→4g(x)=limx→4f(x)7. Therefore, limx→4f(x)=7πlimx→4f(x)=7π.

f(x)={|x|x0forx≠0forx=0 If f is the function defined above, then limx→0f(x) is A −1−1 B 0 C 1 D nonexistent

Answer D Correct. By definition|x|={−xxforx<0forx≥0|x|={−xforx<0xforx≥0.As xx approaches 0 from the left, x<0x<0 and thus |x|=−x|x|=−x. Therefore, the left-hand limit can be evaluated as follows: limx→0−f(x)=limx→0−|x|x=limx→0−(−xx)=limx→0−(−1)=−1limx→0−f(x)=limx→0−|x|x=limx→0−(−xx)=limx→0−(−1)=−1.As xx approaches 0 from the right, x>0x>0 and thus |x|=x|x|=x. Therefore, the right-hand limit can be evaluated as follows: limx→0+f(x)=limx→0+|x|x=limx→0+(xx)=limx→0+1=1limx→0+f(x)=limx→0+|x|x=limx→0+(xx)=limx→0+1=1.Since limx→0−f(x)≠limx→0+f(x)limx→0−f(x)≠limx→0+f(x), it follows that limx→0f(x)limx→0f(x) does not exist.

The function f is defined for all x in the interval 3<x<6. Which of the following statements, if true, implies that limx→5f(x)=12? A There exists a function gg with f(x)≤g(x)f(x)≤g(x) for 3<x<63<x<6, and limx→5g(x)=12limx→5g(x)=12. B There exists a function gg with g(x)≤f(x)g(x)≤f(x) for 3<x<63<x<6, and limx→5g(x)=12limx→5g(x)=12. C There exist functions gg and hh with g(x)≤f(x)≤h(x)g(x)≤f(x)≤h(x) for 3<x<63<x<6, and limx→5g(x)=11limx→5g(x)=11 and limx→5h(x)=13lim⁡x→5h(x)=13. D There exist functions gg and hh with g(x)≤f(x)≤h(x)g(x)≤f(x)≤h(x) for 3<x<63<x<6, and limx→5g(x)=limx→5h(x)=12limx→5g(x)=limx→5h(x)=12.

Answer D Correct. The hypothesis conditions of the squeeze theorem are confirmed, so it may be concluded that limx→5f(x)=12limx→5f(x)=12.

Let f be a function of x. The value of limx→af(x) can be found using the squeeze theorem with the functions g and h. Which of the following could be graphs of f, g, and h ? A B C D

Answer D Correct. The hypothesis conditions of the squeeze theorem are satisfied, since g(x)≤f(x)≤h(x)g(x)≤f(x)≤h(x) on the interval shown and limx→ag(x)=limx→ah(x)limx→ag(x)=limx→ah(x). Therefore, limx→af(x)limx→af(x) can be found using the squeeze theorem with these functions gg and hh.