In the limit #lim_(x->1^-)1/(1-x^2)=oo#, how do you find #delta>0# such that whenever #1-delta<x<1#, #1/(1-x^2)>100#?

Answer 1

See below

It must be #1/(1-x^2)>100#
from here #1>100-100x^2# then transposing terms we have
#100x^2>99# then #x^2>99/10# or equivalent #x>+-sqrt(99/100)#.
we choose positive root because #delta>0#
#x>sqrt99/10#
If we take #delta=sqrt99/10# we are sure that:
first.- #delta <1# second.- if #x in (1-delta,1)# then #1/(1-x^2)>100#
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Answer 2
To find the value of delta, we need to solve the inequality 1/(1-x^2) > 100 for x in the interval (1-delta, 1). First, let's simplify the inequality by multiplying both sides by (1-x^2): 1 > 100(1-x^2) Next, divide both sides by 100: 1/100 > 1-x^2 Rearranging the inequality: x^2 > 1 - 1/100 x^2 > 99/100 Taking the square root of both sides: |x| > √(99/100) Since we are interested in the interval (1-delta, 1), we can ignore the negative values of x. Therefore, we have: x > √(99/100) Simplifying further: x > √99/10 Now, we can choose delta as the difference between 1 and √99/10: delta = 1 - √99/10 This value of delta ensures that whenever 1-delta < x < 1, the inequality 1/(1-x^2) > 100 holds true.
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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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