In the limit #lim (x^2-4x)/(2x)=-2# as #x->4#, how do you find #delta>0# such that whenever #0<abs(x)<delta#, #abs((x^2-4x)/(2x)-(-2))<0.01#?

Answer 1

William Abdalla

Check this question, as:

#lim_(x->4) (x^2-4x)/(2x) = 0#

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Answer 2

Adam Adkins

To find delta>0 such that whenever 00, we can remove the absolute value signs, resulting in delta

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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.

In the limit #lim (x^2-4x)/(2x)=-2# as #x-&gt;4#, how do you find #delta&gt;0# such that whenever #0&lt;abs(x)&lt;delta#, #abs((x^2-4x)/(2x)-(-2))&lt;0.01#?

In the limit #lim (x^2-4x)/(2x)=-2# as #x->4#, how do you find #delta>0# such that whenever #0<abs(x)<delta#, #abs((x^2-4x)/(2x)-(-2))<0.01#?