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How do you find the limit of #(sqrt(x+8)-3)/(x-1)# as #x->1#?

Answer 1

Charles Anctil

#1/6#

You would multiply the terms of the fraction by the factor #sqrt(x+8)+3#:

#lim_(x->1)((sqrt(x+8)-3)* (sqrt(x+8)+3))/((x-1)*(sqrt(x+8)+3))#

#lim_(x->1)(x+8-9)/((x-1)*(sqrt(x+8)+3))#

#lim_(x->1)cancel((x-1))/(cancel((x-1))*(sqrt(x+8)+3))#

#=1/(sqrt(1+8)+3)#

#=1/6#

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Answer 2

Adam Adkins

To find the limit of (sqrt(x+8)-3)/(x-1) as x approaches 1, we can use algebraic manipulation and the concept of limits. By factoring the numerator as (sqrt(x+8)-3) = (sqrt(x+8)-3)(sqrt(x+8)+3), we can simplify the expression to (x+8-9)/(x-1)(sqrt(x+8)+3). Canceling out the (x-1) terms, we are left with (1)/(sqrt(x+8)+3). Substituting x=1 into this expression, we get 1/(sqrt(1+8)+3) = 1/(sqrt(9)+3) = 1/(3+3) = 1/6. Therefore, the limit of (sqrt(x+8)-3)/(x-1) as x approaches 1 is 1/6.

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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.

How do you find the limit of #(sqrt(x+8)-3)/(x-1)# as #x-&gt;1#?

How do you find the limit of #(sqrt(x+8)-3)/(x-1)# as #x->1#?