How do you prove limit of #14-5x=4# as #x-2# using the precise definition of a limit?

Question

How do you prove limit of #14-5x=4# as #x->2# using the precise definition of a limit?

Emilia Aguilar · Accepted Answer

Using #epsilon-delta# definition of limits:

#14-5x=4=>10-5x=0#

Let #f(x)=10-5x#

Thus:

#lim_(x->2)10-5x=0#

Algebraically, this makes sense; yet we want to prove this using the precise definition of a limit.

Since the general formula looks like:

#lim_(x->a)f(x)=L#

This implies that:

#forallepsilon>0, existsdelta>0# such that

#0<|x-a|< delta=>|f(x)-L| < epsilon#

This means that as we pick an interval on the x-axis that is close to #a#there will be an interval on the y-axis that is close to #L#.

So, if we plug in the values we know:

#0<|x-2|< delta =>|(10-5x)-0|< epsilon#

We want to manipulate #|f(x)-L|< epsilon#
so that it can represent #delta# as a function of #epsilon#.

#|10-5x|< epsilon => |-5||x-2|< epsilon=> |x-2|< epsilon/5#

Now they look similar and we can see that #delta = epsilon/5#

This gives us a ratio for when you're given a distance from #L# (or an error tolerance).

So, lets choose #delta = epsilon/5# and plug it back into our delta function.

#0<|x-2|< epsilon/5#

#0<5|x-2|< epsilon#

#0<|5x-10|< epsilon => |f(x)-L|< epsilon#

That concludes our proof.

Christopher Allers · Answer

undefined To prove the limit of 14-5x=4 as x approaches 2 using the precise definition of a limit, we need to show that for any given positive value ε, there exists a positive value δ such that if 0 < |x - 2| < δ, then |(14-5x) - 4| < ε. Let's proceed with the proof: Given ε > 0, we need to find a δ > 0 such that if 0 < |x - 2| < δ, then |(14-5x) - 4| < ε. Starting with |(14-5x) - 4| < ε, we simplify it to |10-5x| < ε. Now, we can manipulate the inequality to isolate x: -ε < 10-5x < ε. Adding 5x to all parts of the inequality, we get: -ε + 5x < 10 < ε + 5x. Subtracting 10 from all parts of the inequality, we have: -ε + 5x - 10 < 0 < ε + 5x - 10. Simplifying further, we get: -ε + 5x - 10 < 0 and 0 < ε + 5x - 10. Adding 10 to all parts of the inequalities, we obtain: -ε + 5x < 10 and 0 < ε + 5x. Dividing both sides of the inequalities by 5, we have: (-ε + 5x)/5 < 10/5 and 0/5 < (ε + 5x)/5. Simplifying, we get: -ε/5 + x < 2 and 0 < ε/5 + x. Subtracting x from both sides of the inequalities, we obtain: -ε/5 + x - x < 2 - x and 0 - x < ε/5 + x - x. Simplifying further, we have: -ε/5 < 2 - x and -x < ε/5. Multiplying both sides of the second inequality by -1 (since x > 0), we get: x > -ε/5. Now, we can choose a suitable value for δ. Let's set δ = min(1, ε/5). If 0 < |x - 2| < δ, then -δ < x - 2 < δ. Substituting the value of δ, we have: -min(1, ε/5) < x - 2 < min(1, ε/5). Since δ ≤ 1, we can further simplify the inequality to: -ε/5 < x - 2 < ε/5. Adding 2 to all parts of the inequality, we obtain: -ε/5 + 2 < x < ε/5 + 2. This shows that if 0 < |x - 2| < δ (where δ = min(1, ε/5)), then -ε/5 + 2 < x < ε/5 + 2. Therefore, we have proven that for any given positive value ε, there exists a positive value δ (specifically, δ = min(1, ε/5)) such that if 0 < |x - 2| < δ, then |(14-5x) - 4| < ε.

How do you prove limit of #14-5x=4# as #x-&gt;2# using the precise definition of a limit?

How do you prove limit of #14-5x=4# as #x->2# using the precise definition of a limit?