Answer:
B
Step-by-step explanation:
(look up limit calculator)
a has a limit
b does not exist
c has a limit
d has a limit
e has a limit
Option D is correct one. Limit doesn't exist for [tex]\lim_{x\to \ 2} \frac{x^{2}-6x +8 }{x-2}[/tex].
The limit of f(x) at x = a is said to be exist if the function approaches the same value from both the sides i.e. LHL =RHL (left hand limit = right hand limit)
According to the question we have,
A. [tex]\lim_{x \to \ 1} \frac{2-x}{x+1}[/tex]
at x =1
LHL = [tex]\lim_{h \to \ 0 } \frac{2- (1-h)}{(1-h)+1}[/tex] = [tex]\frac{1}{2}[/tex]
RHL = [tex]\lim_{h \to \0} \frac{(2-(1+h) }{(1+h)+1)} =\frac{1}{2}[/tex]
Here, RHL = LHL
Hence, limit exist.
B. [tex]\lim_{x \to \ 2} \frac{x^{2} -8}{x-2}[/tex]
at x =2
LHL = [tex]\lim_{h \to \ 0} \frac{(2-h)^{2}-9 }{(2-h)-2}[/tex] =[tex]\lim_{h \to \ 0} \frac{4+h^{2} -9-4h}{2-h-2}[/tex] = [tex]\lim_{h \to \ 0 }\frac{2h-4}{-1} =\frac{-4}{-1} =4[/tex]
RHL = [tex]\lim_{h \to \ 0} \frac{(2+h)^{2} -9}{(2+h)-2}[/tex] = [tex]\lim_{h \to \ 0} \frac{4h +4 +h^{2}-9 }{h} = \lim_{h \to \ 0} \frac{2h+4}{1}[/tex]=4
Here, RHL = LHL
Hence, the limit exist.
C. [tex]\lim_{x \to \ 1} (x^{2} -2x-5)[/tex]
at x =1
LHL = [tex]\lim_{h \to \ 0} (( 1-h)^{2} -2(1-h)-5)= \lim_{h \to \ 0} (1 -2-5)=-6[/tex]
RHL = [tex]\lim_{h \to \ 0} ((1+h)^{2} -2(1+h)-5) = (1-2-5)=-6[/tex]
Here, LHL = RHL = -6
Hence, the limits exist.
D. [tex]\lim_{x \to \ 2} \frac{x^{2}-6x + 8 }{x - 2}[/tex]
at x =2
LHL = [tex]\lim_{h \to \0} \frac{(2-h)^{2} -(2-h)+8}{(2-h)-2} = \lim_{h \to \ 0} \frac{4+h^{2}-4h-2+h+8 }{(2-h)-2}[/tex] =[tex]\lim_{h \to \ 0} \frac{2h-4+1}{1} =-3[/tex]
RHL = [tex]\lim_{h \to \ 0} \frac{(2+h)^{2} -(2+h) +8}{(2+h)-2} = \lim_{h \to \ 0} \frac{4+h^{2}+4h -2-h+8}{h} = \lim_{h \to \ 0} \frac{2h+4-1}{1} =3[/tex]
Here, RHL���LHL
Hence, the limit doesn't exist.
E. [tex]\lim_{x \to \ 3} (x^{2} -3x-1)[/tex]
at x = 3
LHL = [tex]\lim_{h \to \ 0} ((3-h)^{2} -3(3-h)-1)[/tex] = [tex]9-9-1=-1[/tex]
RHL= [tex]\lim_{h \to \ 0} ((3+h)^{2} -3(3+h)-1)= 9-9-1 =-1[/tex]
Here, RHL = LHL
Hence, the limit exist.
Learn more about limit here:
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