# 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#?

Please see below.

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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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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.

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.

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.

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