Question regarding limits?

Question

I would like to see if these are right.

for the lim g(x), where g(x) is sqrt(1-x) when x is less than or equal to 1

x + 1 when x is greater than 1

Does the limit as x approaches 1 from the right exist? Yes, the limit exists because when x approaches one it's zero?

Also, how do would you solve lim (sqrt (2x+1) -sqrt(3)) / x-1 algebraically? I know how to graphically, but how would I rationalize that?

Answer 1

First question:

lim g(x) as x approaches 1 from the left

= lim x→1-  Ã¢ÂˆÂš(1-x)

= √(1-1)

= 0

lim g(x) as x approaches 1 from the right

= lim x→1+ (x+1)

= (1+1)

=2

Therefore limits DO exist as x approaches 1 from the left and as x approaches 1 from the right BUT NO limit exists at 1, since these two limits are different.

Second question

lim   [(√(2x+1) - √3)] / (x-1)

x→1

= lim [(√(2x+1) - √3) (√(2x+1) + √3)] / [(x-1) (√(2x+1) + √3)]

= lim [(2x+1)-(3)] / (x-1)(√(2x+1) + √3)

= lim [2(x-1)] / [(x-1)(√(2x+1) + √3)]

Simplifying, we get

= lim 2 / (√(2x+1) + √3)  At this point we can now substitute x with 1

= 2 / (√3 + √3)

= 2 /(2√3)  

= 1/√3

To rationalize the denominator, multiply both the numerator and denominator by √3

= √3/(√3√3)

=√3/3

P.S. I tried to answer this question before, but I was told it had been deleted when I tried to post it. I got a different answer then, and was wondering if the problem had been written down incorrectly. I guess it had, since this time I am getting the answer you wanted the first time :)

Answer 2

lim of g(x) as x approaches 1 from the right is 2 and not zero.

lim x → 1 [√(2x+1) - √3] / (x - 1)

= lim x → 1 [√(2x+1) - √3]*[√(2x+1) + √3] / (x - 1)*[√(2x+1) + √3]

= lim x → 1 [2(x-1)] / (x - 1)*[√(2x+1) + √3]

= 2 lim x → 1 1/[√(2x+1) + √3]

= 1/√3.

Answer 3

(sqrt (2x+1) -sqrt(3)) / x-1 = (sqrt (2x+1) -sqrt(3)) (x+1) (sqrt (2x+1) + sqrt(3)) / (x-1)(x+1)  (sqrt (2x+1) + sqrt(3))  = (2x+1) -(3)) (x+1)/ (x²-1) (sqrt (2x+1) + sqrt(3))

wait, the limit when x approaches what? 1?