What is the limit of the function [sqrt(x-1)] as x approaches 1?

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In summary, the limit of sqrt(x-1) as x approaches 1 is undefined. However, certain computer programs may give a value of 0 due to working in the complex plane. It is important to clarify with your teacher or instructor on how to approach such problems in assignments.
  • #1
khurram usman
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limit x-->1 of [sqrt(x-1)]

i know that for limit to be defined the left and right handed limits should be equal.
in the above problem, the left handed limit is undefined because it is the square rot of a negative no. but i checked this question on alphawolfram.com and also on MATHEMATICA . both the programs were giving limit as 0
so i supposed i was missing something but when i submitted my answer as 'undefined' it was marked as correct and i was given full credit
so please everyone clarify this problem
and also consider this one: lim x-->1 of sqrt[1-x]
in this case the right hand liit is not defined but again mathematica gave the same answer '0'
 
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  • #2


khurram usman said:
i know that for limit to be defined the left and right handed limits should be equal.
in the above problem, the left handed limit is undefined because it is the square rot of a negative no. but i checked this question on alphawolfram.com and also on MATHEMATICA . both the programs were giving limit as 0
so i supposed i was missing something but when i submitted my answer as 'undefined' it was marked as correct and i was given full credit
so please everyone clarify this problem
and also consider this one: lim x-->1 of sqrt[1-x]
in this case the right hand liit is not defined but again mathematica gave the same answer '0'
You are correct in saying that for [itex]f:\mathbb{R}^+\mapsto\mathbb{R}^+[/itex], [itex]f:x\mapsto\sqrt{x}[/itex], the left-hand limit does not exist. However, Mathematica and Wolfram Alpha (and Maple, I would imagine as well as most CAS) will quite happily work in the complex plane, where the limit does exist and is indeed zero.
 
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  • #3


yes you are right there.
on the graph one of the lines was marked as real and other as imaginary.
so it means as far as i have learned, my answer was correct. the limit for both the functions is un defined
 
  • #4


khurram usman said:
so it means as far as i have learned, my answer was correct. the limit for both the functions is un defined
Both functions, there is only one function: the square root. The two lines on the graph indicate the values of the function (both the real and imaginary parts).
 
  • #5


ok...but i have not been taught any complex valued functions yet.
so as far as my assingnment is concernerd i should submit this as undefined.
thanks a lot
 
  • #6


khurram usman said:
so as far as my assingnment is concernerd i should submit this as undefined.
It would seem so, yes.
khurram usman said:
thanks a lot
No problem.
 

1. What is the value of the limit x→1 of √(x-1)?

The value of the limit x→1 of √(x-1) is 0.

2. How do you evaluate the limit x→1 of √(x-1)?

To evaluate the limit x→1 of √(x-1), we can use the substitution method by plugging in 1 for x. This results in √(1-1) which simplifies to √0 = 0.

3. Is the limit x→1 of √(x-1) continuous?

Yes, the limit x→1 of √(x-1) is continuous. This is because the limit exists at x=1 and the function is defined at x=1.

4. What is the domain of the function √(x-1)?

The domain of the function √(x-1) is all real numbers greater than or equal to 1, since the square root of a negative number is undefined.

5. How does the graph of √(x-1) look like near x=1?

The graph of √(x-1) near x=1 is a smooth curve approaching the point (1,0) from the left side of the graph. It does not intersect with the line x=1, as the function is not defined at x=1.

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