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?

1 Answer
Jul 30, 2017

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

Explanation:

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.