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In the limit #lim sqrt(6-3x)=0# as #x->2^-#, how do you find #delta>0# such that whenever #2-delta<x<2#, #sqrt(6-3x)<0.01#?

Answer 1

Scarlett Allison

Please see below.

Figure out how close #x# must be to #2#. (That is #delta#.)

#sqrt(6-3x) < 1/100#

Square both sides: #6-3x < 1/10^4#

Factor the #3# on the left #3(2-x) < 1/10^4#

So, #2-x < 3/10^4 = 0.0003#

And we need

#2-0.0003 < x#

#delta = 0.0003#

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Answer 2

Emma Aldridge

To find the value of delta, we need to solve the inequality sqrt(6-3x) < 0.01.

First, let's solve the inequality without the square root: 6-3x < 0.01^2.

Simplifying, we have 6-3x < 0.0001.

Next, isolate x by subtracting 6 from both sides: -3x < -5.9999.

Divide both sides by -3, remembering to reverse the inequality sign: x > 1.9999.

Therefore, for sqrt(6-3x) < 0.01 to hold true, we need x to be greater than 1.9999.

Since we are looking for a delta such that whenever 2-delta < x < 2, sqrt(6-3x) < 0.01, we can set delta = 0.0001.

Thus, whenever 2-0.0001 < x < 2, sqrt(6-3x) < 0.01 will be satisfied.

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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 sqrt(6-3x)=0# as #x-&gt;2^-#, how do you find #delta&gt;0# such that whenever #2-delta&lt;x&lt;2#, #sqrt(6-3x)&lt;0.01#?

In the limit #lim sqrt(6-3x)=0# as #x->2^-#, how do you find #delta>0# such that whenever #2-delta<x<2#, #sqrt(6-3x)<0.01#?