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How do you find the limit of #abs(8x-56)/ (x-7)# as #x->7^-#?

Answer 1

You need to consider exclusively values of #x < 7#

Since #x -> 7^-# we should consider only values of #x < 7#. In that case we have #8x < 56#, and therefore #8x - 56 < 0#. Hence:

#|8x - 56| = -8x + 56#, and so the expression becomes:

#(-8x+56)/ (x-7)= - 8 (x -7)/(x-7)#. We can simplify #(x-7)# since #x != 7#, so we have:

#lim_(x-> 7^-)(-8x+56)/ (x-7)= lim_(x-> 7^-)- 8 (x -7)/(x-7)= -8#

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Answer 2

Christopher Agresta

To find the limit of abs(8x-56)/ (x-7) as x approaches 7 from the left (x->7^-), we can substitute the value of x into the expression. However, since the denominator becomes zero at x=7, we need to simplify the expression before substituting.

First, we can factor out 8 from the numerator: abs(8(x-7))/ (x-7).

Next, we can cancel out the common factor of (x-7) in the numerator and denominator: abs(8).

Therefore, the limit of abs(8x-56)/ (x-7) as x approaches 7 from the left is simply 8.

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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 #abs(8x-56)/ (x-7)# as #x-&gt;7^-#?

How do you find the limit of #abs(8x-56)/ (x-7)# as #x->7^-#?