Guess the value of the limit ( it it exists) by evaluationg the funciton at the given numbers ( correct to six decimal places) a) lim x^(2)-2x / x^(2)-x-2, x=2.5,2.1,2.05,2.01,2.005,2.001,1.9,1.95,199,1.995,1.999 x rightarrow 2 b) lim x^(2)-2x/ x^(2)-x-2, x=0, -0.5,-0.9, -0.95, -0.99, -0.999, -2, -1.5, -1.1, -1.01, -1.001 x rightarrow -1

crazygbyo

crazygbyo

Answered question

2022-08-05

Guess the value of the limit ( it it exists) by evaluationg the funciton at the given numbers ( correct to six decimal places)
a) lim x 2 2 x / x 2 x 2 , x = 2.5 , 2.1 , 2.05 , 2.01 , 2.005 , 2.001 , 1.9 , 1.95 , 199 , 1.995 , 1.999
x 2
b) lim x 2 2 x / x 2 x 2 , x = 0 , 0.5 , 0.9 , 0.95 , 0.99 , 0.999 , 2 , 1.5 , 1.1 , 1.01 , 1.001
x 1

Answer & Explanation

agergadas3b

agergadas3b

Beginner2022-08-06Added 16 answers

a) lim x 2 2 x / x 2 x 2 , x = 2.5 , 2.1 , 2.05 , 2.01 , 2.005 , 2.001 , 1.9 , 1.95 , 199 , 1.995 , 1.999
x 2
f ( x ) = ( x 2 2 x ) / ( x 2 x 2 )
f(2.5)=0.714286
f(2.1)=0.677419
f(2.05)=0.672131
f(2.01)=0.667774
f(2.005)=0.667221
f(2.001)=0.666778
f(1.9)=0.655172
f(1.95)=0.661017
f(1.99)=0.665552
f(1.995)=0.66611
f(1.999)=0.666556
lim x 2 2 x / x 2 x 2 = 2 / 3
x 2
b) lim x 2 2 x / x 2 x 2 , x = 0 , 0.5 , 0.9 , 0.95 , 0.99 , 0.999 , 2 , 1.5 , 1.1 , 1.01 , 1.001
x 1
f ( x ) = ( x 2 2 x ) / ( x 2 x 2 )
f(0)=0
f(-0.5)=-1
f(-0.9)=-9
f(-0.95)=-19
f(-0.99)=-99
f(-0.999)=-999
f(-2)=2
f(-1.5)=3
f(-1.1)=11
f(-1.01)=101
f(-1.001)=1001

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