In the limit #lim sqrtx=0# as #x->0^+#, how do you find #delta>0# such that whenever #0<x<delta#, #sqrtx<0.01#?

Answer 1
As #y=sqrtx# is a strictly increasing function:
#0 < x < epsilon^2 <=> 0 < sqrt(x) < sqrt(epsilon^2)#
So, for any #epsilon > 0#, if we take #delta_epsilon = epsilon^2# we have:
#0 < x < delta_epsilon <=> 0 < sqrt(x) < epsilon#
In particular for #epsilon =0.01# we can choose #delta_epsilon <= 0.0001#
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Answer 2

To find a suitable delta value, we can start by setting the given condition: sqrt(x) < 0.01.

Next, we can square both sides of the inequality to eliminate the square root: x < 0.0001.

Now, we need to find a delta value such that whenever 0 < x < delta, the condition x < 0.0001 holds.

Since we want to find a delta value for the limit as x approaches 0 from the positive side (x -> 0+), we can choose delta = 0.0001.

Therefore, whenever 0 < x < 0.0001, the condition sqrt(x) < 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.

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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