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In the limit #lim 5/sqrt(2-x)=oo# as #x->2^-#, how do you find #delta>0# such that whenever #2-delta<x<2#, #5/sqrt(2-x)>100#?

Answer 1

Evelyn Agard

Solve the inequality for #x#, then find #delta#.

Since #sqrt(2-x)# is positive, we can mu;tiply without changing the inequality, so we want

#5/100 > sqrt(2-x)#

#sqrt(2-x) < 0.05#

Both are positive and if #a > b# and both are positive, then #a^2 > b^2#. So we want

#2-x < 0.0025# and

#2-0.0025 < x#

#delta = 0.0025#

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Answer 2

Adam Adkins

To find delta>0 such that whenever 2-delta100, we can start by solving the inequality 5/sqrt(2-x)>100. First, we can multiply both sides of the inequality by sqrt(2-x) to get rid of the denominator: 5 > 100*sqrt(2-x). Next, we can divide both sides of the inequality by 100: 5/100 > sqrt(2-x). Simplifying further, we have: 1/20 > sqrt(2-x). To isolate the square root, we can square both sides of the inequality: (1/20)^2 > 2-x. Simplifying, we get: 1/400 > 2-x. Rearranging the inequality, we have: x > 2 - 1/400. Therefore, we can choose delta = 1/400.

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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 5/sqrt(2-x)=oo# as #x-&gt;2^-#, how do you find #delta&gt;0# such that whenever #2-delta&lt;x&lt;2#, #5/sqrt(2-x)&gt;100#?

In the limit #lim 5/sqrt(2-x)=oo# as #x->2^-#, how do you find #delta>0# such that whenever #2-delta<x<2#, #5/sqrt(2-x)>100#?