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How do you find the limit of #((root3 x) -1)/((sqrt x) -1)# as x approaches 1?

Answer 1

Emma Aldridge

Rationalize and reduce.

#(sqrtx-1)(sqrtx+1) = x-1# #" "# (use #a^2-b^2=(a-b)(a+b)#)

and

#(root3x-1)(rootrx^2+root3x+1) = (x-1)# (use #a^3-b^3 = (a-b)(a^2+ab+b^2)#)

So,

#((root3 x) -1)/((sqrt x) -1) = ((x-1)(sqrtx+1))/((x-1)(rootrx^2+root3x+1))#

# = (sqrtx+1)/(rootrx^2+root3x+1)#

Taking the limit as #xrarr1# yields #2/3#.

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Answer 2

Charles Anctil

To find the limit of ((√3x) - 1)/((√x) - 1) as x approaches 1, we can use algebraic manipulation. By multiplying the numerator and denominator by the conjugate of (√x) - 1, we can simplify the expression. After simplification, we can substitute x = 1 into the expression to find the limit. The limit of ((√3x) - 1)/((√x) - 1) as x approaches 1 is 2√3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.