18.)
}+\frac{a}{a^{x}(a-1)})^{\frac{1}{x}})
Term inside the outermost brackets is between -1 and 1 when x is beyond a certain value, provided that a is not between -1 and 1. Therefore that term raised to some number is also between -1 and 1.
}+\frac{a}{a^{x}(a-1)})^{\frac{1}{x}} < 1)
}+\frac{a}{a^{x}(a-1)})^{\frac{1}{x}} < \frac{a}{x})
Now take limit as x approaches infinity of all sides (aka sandwhich theorem or squeeze theorem)
Hence the thing equals 0 for all values
not in (-1,1].
Btw: the above is for a>1. For a<-1 we need to reverse the inequality, which still gives the same answer.
For value a=0. The limit is obviously 0.
For values in (0,1):Looking at the original expression, it is obvious that the
term can be made as small as one wishes by making x large enough. At some value of x, the expression is some value M. Hence by increasing the value of x the modulus of the expression becomes less than the modulus of M. The expression is negative, so the expression becomes greater than M (less negative).
Hence we know that for all values of x beyond some number:
<0)
^{\frac{1}{x}}<0)
Now take the limit as x approaches infinity of these terms and u find that the limit of the lower bound is zero since the
gets smaller in modulus as x approaches infinity(in fact it approaches -1), while the
bit appraoches 0. And so using the product property the limit is 0 and so the limit asked in the question is also 0 for a in (0,1).
:)
the term
is between 0 and 1. The modulus of the term
can be made as small as we like. Hence when x is beyond some value the term
will always be between 0 and 1. Using sandwhich theorem again gives the required result.
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