In the limit #lim_(t to oo) 1/(1+4t)=0# , how do you find #B0# such that whenever #tB#, #1/(1+4t)

Question

In the limit #lim_(t to oo) 1/(1+4t)=0# , how do you find #B>0# such that whenever #t>B#, #1/(1+4t)<0.01#?

Elijah Abram · Accepted Answer

Just solve the inequality #1/(1+4t)<0.01# to get #t>24.75# (so #B=24.75#).

The inequality #1/(1+4t)<0.01=1/100# is equivalent to #1+4t>100# (flip the fractions and reverse the direction of the inequality). This, in turn, is equivalent to #4t>99# and equivalent to #t>99/4=24.75#. For a function #f# defined for all #t>0#, the definition of #lim_{t -> infty}f(t)=L# can be stated like this: for all #epsilon>0#, there exists a number #B# such that #|f(t)-L| < epsilon# for all #t > B#. (The value of #f(t)# can be made to be a distance from #L# less than any #epsilon>0# if the input #t# is sufficiently large...#B# is the measure of "sufficiently large"). The point of this problem for #f(t)=1/(1+4t)# and #L=0# is to find the corresponding value of #B# when #epsilon=0.01#. This does not prove that #lim_{t -> infty}f(t)=0#, but is equivalent to the kind of algebra you would need to do to help you do a proof. Assuming #t>0#, we have #1/(1+4t) < epsilon \leftrightarrow 1+4t>1/epsilon \leftrightarrow t > (1/epsilon - 1)/4=(1-epsilon)/(4epsilon)#, we can do the proof as follows: Proof: Let #epsilon > 0# be given and let #B=(1-epsilon)/(4epsilon)#. Suppose #t > 0# also satisfies #t>B#. It follows that #1+4t>1/epsilon# and therefore #0 < f(t) =1/(1+4t) < epsilon# (the first inequality is obvious here since #t>0#). Therefore, #|f(t) - 0| < epsilon#. This proves that #lim_{t -> infty}f(t)=0#.

Luke Alvarez · Answer

undefined To find B>0 such that whenever t>B, 1/(1+4t)<0.01, we can set up an inequality and solve for B. Starting with the given limit, lim_(t to oo) 1/(1+4t)=0, we can rewrite it as 1/(1+4t) - 0 < ε, where ε is a positive number (in this case, ε=0.01). Now, we can solve the inequality 1/(1+4t) < ε, which is equivalent to 1/(1+4t) < 0.01. Multiplying both sides by (1+4t), we get 1 < 0.01(1+4t). Expanding the right side, we have 1 < 0.01 + 0.04t. Subtracting 0.01 from both sides, we obtain 0.99 < 0.04t. Dividing both sides by 0.04, we get 24.75 < t. Therefore, B>0 can be chosen as B=24.75, and whenever t>B, 1/(1+4t)<0.01.

In the limit #lim_(t to oo) 1/(1+4t)=0# , how do you find #B&gt;0# such that whenever #t&gt;B#, #1/(1+4t)&lt;0.01#?

In the limit #lim_(t to oo) 1/(1+4t)=0# , how do you find #B>0# such that whenever #t>B#, #1/(1+4t)<0.01#?