limits question

**alpha** · January 18th 2010, 12:25 PM

explain why the limit as x tends to positive 0 of 2^(1/x) = infinity but the limit as x tends to negative 0 of 2^(1/x)=0.

Thanks.

**General** · January 18th 2010, 12:27 PM

Originally Posted by alpha

explain why the limit as x tends to positive 0 of 2^(1/x) = infinity but the limit as x tends to negative 0 of 2^(1/x)=0.

i can see it on the graph that i have sketched but i can't explain it.

Thanks.

The graph itself is the best answer for your question.

**Raoh** · January 18th 2010, 12:33 PM

Originally Posted by alpha

explain why the limit as x tends to positive 0 of 2^(1/x) = infinity but the limit as x tends to negative 0 of 2^(1/x)=0.

Thanks.

hi
Hint,
$2^{\frac{1}{x}}=\exp(\log(2^{\frac{1}{x}}))$

**alpha** · January 18th 2010, 12:40 PM

Originally Posted by Raoh

hi
Hint,
$2^{\frac{1}{x}}=\exp(\log(2^{\frac{1}{x}}))$

i can see it's the same graph but what is the explanation for this.

thanks.

**Raoh** · January 18th 2010, 12:46 PM

Originally Posted by alpha

i can see it's the same graph but what is the explanation for this.

thanks.

$\lim_{x\to 0^{+}}\log(2^{\frac{1}{x}})=\lim_{x\to 0^{+}}\frac{1}{x}\log(2)=\infty$
$\lim_{x\to 0^{-}}\log(2^{\frac{1}{x}})=\lim_{x\to 0^{-}}\frac{1}{x}\log(2)=-\infty$
Remember,
$e^{\infty }=\infty$ and $e^{-\infty }=0$

**Archie Meade** · January 18th 2010, 01:04 PM

$x=1,\ 2^{\frac{1}{x}}=2^1=2$

$x=0.5,\ 2^{\frac{2(0.5)}{0.5}}=2^2$

$x=0.1,\ 2^{\frac{10(0.1)}{0.1}}=2^{10}$

$As\ x\ goes\ to\ zero\ from\ above,\ f(x)\ goes\ to\ infinity.$

$x=-1,\ f(-1)=2^{-1}=\frac{1}{2}$

$x=-0.5,\ f(-0.5)=2^{-2}=\frac{1}{2^2}$

$x=-0.1,\ f(-0.1)=2^{-10}=\frac{1}{2^{10}}$

$As\ x\ goes\ to\ zero\ from\ below,\ f(x)\ goes\ to\ zero.$

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