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Re: Math Quiz 7
I'm perhaps a little late, but it's not actually necessary to use l'Hospital's: we can just use the limit definition of the exponential function.
We have that (1 - a/n)n converges to e-a. Thus, the sequence
a_k = (1-3/(k+3))2k+6 = ((1-3/(k+3))k+3)2
will converge to (e-3)2 = e-6. The sequence of terms we have is
b_k= a_k / (1-3/(k+3))6
and (1-3/(k+3))6 goes to 1 as k becomes large, so in the limit b_k converges to e-6/1 = e-6.
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