Facing difficulty finding limit \(\displaystyle\lim_{{{x}\to\infty}}{\left({\frac{{{x}}}{{{x}-{1}}}}\right)}^{{{2}{x}+{1}}}\)

Destinee Hensley 2022-04-04 Answered
Facing difficulty finding limit
limx(xx1)2x+1
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Answers (2)

membatas0v2v
Answered 2022-04-05 Author has 19 answers
If you know that
limx(1+ax)x=ea
so that
limx(11x)x=e1
then you can try to rewrite your limit into something involving this limit.
So try rewriting it; perhaps as a product,
(xx1)2x+1=((xx1)x)2(xx1)
=(1(x1x)x)2(xx1)
=(1(11x)x)2(xx1)
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clarkchica44klt
Answered 2022-04-06 Author has 17 answers
limx(xx1)2x+1=limx(x1+1x1)2x+1
=limx(1+1x1)2x+1=limx(1+1x1)(x1)2x+1x1
=limx(1+1x1)(x1)2x+1x1
=elimx2x+1x1=e2
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