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How do you find the limit of #sqrt(x-4)/(x-16)# as x approaches 16?

Answer 1

Isaiah Allen

#1/8#

We can factor #x-16# as a difference of squares.

Since differences of squares take the pattern #a^2-b^2=(a+b)(a-b)#, we see that in #x-16#,

#a^2=x" "=>" "a=sqrtx# #b^2=16" "=>" "b=4#

Yielding:

#x-16=(sqrtx+4)(sqrtx-4)#

The problem then becomes:

#lim_(xrarr16)(sqrtx-4)/(x-16)=lim_(xrarr16)(sqrtx-4)/((sqrtx+4)(sqrtx-4))=lim_(xrarr16)1/(sqrtx+4)#

Evaluate the limit by plugging in #16#:

#lim_(xrarr16)1/(sqrtx+4)=1/(sqrt16+4)=1/(4+4)=1/8#

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Answer 2

Ezra Adams

To find the limit of sqrt(x-4)/(x-16) as x approaches 16, we can use algebraic manipulation and the concept of limits. By factoring the denominator as (x-16), we can cancel out the common factor of (x-16) in both the numerator and denominator. This simplifies the expression to sqrt(x-4)/1, which is equivalent to sqrt(x-4).

Now, as x approaches 16, the expression sqrt(x-4) approaches sqrt(16-4) = sqrt(12) = 2√3. Therefore, the limit of sqrt(x-4)/(x-16) as x approaches 16 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.