- #1

- 9

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Find limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

for (1) 0<a<1

(2) a>1

- Thread starter chiakimaron
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- #1

- 9

- 0

Find limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

for (1) 0<a<1

(2) a>1

- #2

malawi_glenn

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what have you tried? If you are not showing your effort, then you can't get any help

- #3

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let y=〔 (a^x - 1 )/(a - 1) 〕^1/x

Iny=1/x In(a^x - 1 )/(a - 1)

then I can't do it

Iny=1/x In(a^x - 1 )/(a - 1)

then I can't do it

- #4

malawi_glenn

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start with 0<a<1 then what happens?

Last edited:

- #5

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i really don't know

- #6

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limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

= [(-1)/(a-1)]^0 = 1

is it correct?

= [(-1)/(a-1)]^0 = 1

is it correct?

- #7

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Keep going! Let [tex]x\to\infty[/tex]let y=〔 (a^x - 1 )/(a - 1) 〕^1/x

Iny=1/x In(a^x - 1 )/(a - 1)

then I can't do it

[tex]\lim_{x\to\infty}\frac{\ln\left(\frac{a^x-1}{a-1}\right)}{x}=\frac{\lim_{x\to\infty}\ln\left(\frac{a^x-1}{a-1}\right)}{\lim_{x\to\infty}x}[/tex]

and remember ln(-1/(a-1)) is some constant.

- #8

malawi_glenn

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yes! :-)limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

= [(-1)/(a-1)]^0 = 1

is it correct?

And what if a>1 ?

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