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In the limit #lim 1/(4-t)=-oo# as #t->4^+#, how do you find #delta>0# such that whenever #4<t<t+delta#, #1/(4-t)<-100#?

Answer 1

Charles Anctil

#delta = 1/100#

For any #delta > 0# and #4 < t < t+delta# we have that:

#4-t < 0#

therefore:

#1/(4-t) < -100 => 1 > -100(4-t)#

#1 > -400 +100t#

#t < 401/100#

#t < 4 +1/100#

which shows that we can choose #delta = 1/100#

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Answer 2

William Abdalla

To find delta>0 such that whenever 4<t<t+delta, 1/(4-t)<-100, we can start by solving the inequality 1/(4-t)<-100.

First, we can multiply both sides of the inequality by (4-t) to get rid of the denominator. This gives us 1 < -100(4-t).

Next, we can simplify the inequality by distributing -100 on the right side: 1 < -400 + 100t.

Then, we can add 400 to both sides of the inequality: 401 < 100t.

Dividing both sides by 100, we get: 4.01 < t.

Therefore, we have found that whenever 4<t<4.01, the inequality 1/(4-t)<-100 holds true. Hence, we can choose delta=0.01 as a suitable value.

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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 1/(4-t)=-oo# as #t-&gt;4^+#, how do you find #delta&gt;0# such that whenever #4&lt;t&lt;t+delta#, #1/(4-t)&lt;-100#?

In the limit #lim 1/(4-t)=-oo# as #t->4^+#, how do you find #delta>0# such that whenever #4<t<t+delta#, #1/(4-t)<-100#?