Power Series and Limits

Key Questions

  • To be honest, I would not use power series on this one since this is a perfect problem to demonstrate the application of Squeeze Theorem. Here is how:

    We know

    #-1 le sinx le 1#

    #Rightarrow -3 le 3sinx le 3#

    #Rightarrow -3/e^x le {3sinx}/e^x le 3/e^x#.

    Since

    #lim_{x to infty}(-3/e^x)=-3/infty=0#

    and

    #lim_{x to infty}3/e^x=3/infty=0#,

    we conclude that

    #lim_{x to infty}{3sinx}/e^x=0#

    by Squeeze Theorem.


    I hope that this was helpful.

  • Here is a simple application of a power series in evaluating a limit.

    #lim_{x to 0}{sinx]/x#

    by replacing #sinx# by its Maclaurin series.

    #=lim_{x to 0}{x-x^3/{3!}+x^5/{5!}-x^7/{7!}+cdots}/{x}#

    by distributing the division to each term,

    #=lim_{x to 0}(1-x^2/{3!}+x^4/{5!}-x^6/{7!}+cdots)#

    by sending #x# to zero,

    #=1-0+0-0+cdots#

    since all but the first term are zero,

    #=1#


    I hope that this was helpful.

Questions