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In the limit #lim 10x=40# as #x->4#, how do you find #delta>0# such that whenever #0<abs(x-4)<delta#, #abs(10x-40)<0.01#?

Answer 1

Elijah Abram

#0 lt delta le 0.001#.

#|10x-40| lt 0.01,#

#iff |10(x-4)| lt 0.01,#

#iff |10||x-4| lt 0.01,...[because, |ab|=|a||b|]#

#iff 10|x-4| lt 0.01,#

#iff |x-4| lt 0.01/10=0.001,#

Comparing with, #|x-4| lt delta,# we find that,

#0 lt delta le 0.001#.

Observe that, in the event that, #0 lt delta le 0.001,#

#0 lt |x-4| lt delta le 0.001 rArr |10x-40| lt 0.01#.

Thus, the reqd. #delta" is given by, "0 lt delta le 0.001#.

Enjoy Maths.!

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Answer 2

Ezra Adams

To find the value of delta, we can start by manipulating the given expression abs(10x-40)<0.01.

First, we can simplify the expression by dividing both sides by 10, which gives us abs(x-4)<0.001.

Now, we need to find a value for delta such that whenever 0<abs(x-4)<delta, the inequality abs(x-4)<0.001 holds true.

Since the absolute value of a number is always positive, we can rewrite the inequality as -0.001<(x-4)<0.001.

To ensure this inequality holds true, we can set delta=0.001.

Therefore, whenever 0<abs(x-4)<0.001, the inequality abs(10x-40)<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.

In the limit #lim 10x=40# as #x-&gt;4#, how do you find #delta&gt;0# such that whenever #0&lt;abs(x-4)&lt;delta#, #abs(10x-40)&lt;0.01#?

In the limit #lim 10x=40# as #x->4#, how do you find #delta>0# such that whenever #0<abs(x-4)<delta#, #abs(10x-40)<0.01#?