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In the limit #lim 1/(10t)=oo# as #t->0^+#, how do you find #delta>0# such that whenever #0<t<delta#, #1/(10t)>100#?

Answer 1

See process below

Let be a #t>0#

We want #100<1/(10t)# that it's the same #1000t<1# or #t<1/1000#

Choosing #delta=1/1000>0# we get our goal

By one hand #t < delta#, by other hand #delta>0# and finally choosing #t < delta# (lets say #t=1/10000#) we get the result

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Answer 2

Elijah Abram

To find delta>0 such that whenever 0100, we can start by setting 1/(10t) greater than 100 and solving for t. 1/(10t) > 100 Multiplying both sides by 10t: 1 > 1000t Dividing both sides by 1000: 1/1000 > t Therefore, we have found that t should be less than 1/1000. To find delta, we can simply choose any positive value smaller than 1/1000. For example, we can choose delta = 1/2000. So, whenever 0100.

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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/(10t)=oo# as #t-&gt;0^+#, how do you find #delta&gt;0# such that whenever #0&lt;t&lt;delta#, #1/(10t)&gt;100#?

In the limit #lim 1/(10t)=oo# as #t->0^+#, how do you find #delta>0# such that whenever #0<t<delta#, #1/(10t)>100#?