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I need to find the following limit \begin{align} \lim_{x \to 0} \left\{f(x) = 2^{\frac{1}{x}}\frac{x-1}{x-2}\right\} \end{align} After plotting this thing I got $\lim_{x\to 0^-} f(x) = 0$ and $\lim_{x\to 0^+} f(x) = \infty$. I am not too sure, however, how one would show it analytically.

clueless
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4 Answers4

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The factor $(x-1)/(x-2)$ is well behaved around zero, the problem is $2^{1/x}$, note that

$$ 2^{1/x} = e^{\ln 2^{1/x}} = e^{\ln 2/x} $$

and remember that

$$ \lim_{x\to +\infty} e^x = +\infty ~~~~\mbox{and}~~~ \lim_{x\to -\infty} e^x = 0 $$

With these two things together

$$ \lim _{x\to 0^+}2^{1/x} = \lim _{x\to 0^+} e^{\ln2 / x} = \lim _{y\to +\infty} e^{y\ln 2 } = +\infty $$

whereas

$$ \lim _{x\to 0^-}2^{1/x} = \lim _{x\to 0^-} e^{\ln2 / x} = \lim _{y\to -\infty} e^{y\ln 2 } = 0 $$

caverac
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Change variable as $x\rightarrow\frac{1}{x}$ and you will get $$ \lim_{x\rightarrow\infty}2^x\frac{1-x}{1-2x} $$ and you should see the result.

Jon
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  • Should you not distinguish between the left and right limit, i.e. a positive and a negative value for x? The former results in $+\infty$, the latter in $0$. – jvdhooft May 18 '17 at 11:58
  • Yes, you get different values running from left or right remembering that $2^x$ run to zero for $x\rightarrow -\infty$ and to infinity for $x\rightarrow\infty$. – Jon May 18 '17 at 12:01
  • I have not got into deeper because this approach has been exploited thoroughly in another answer. – Jon May 18 '17 at 12:06
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$$\lim_{x \to 0}2^{\frac{1}{x}}\times \underbrace{\left( \frac{x-1}{x-2}\right)}_{\text{No indeterminacy here}}=\lim_{x \to 0}\frac{1}{2}2^{\frac{1}{x}} $$

Now put for $x \to 0^+$ i.e. $x=0+h$

$$\lim_{x \to 0^+}\frac{1}{2}2^{\frac{1}{h}} \longrightarrow 2^{\infty} \longrightarrow \infty$$

Now put for $x \to 0^-$ ,$x=0-h$

$$\lim_{x \to 0^-}\frac{1}{2}2^{\frac{1}{-h}}\longrightarrow 2^{-\infty} \longrightarrow 0$$

Jaideep Khare
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You can split the calculation:

$$\lim_{x\to 0^+}2^{1/x}=\infty$$

$$\lim_{x\to 0^-}2^{1/x}=0$$

so the above limit doesn't exist.

Also,

$$\lim_{x\to 0}\frac{x-1}{x-2}=\frac{1}{2}$$

and then (using that $\lim_{x\to x_0}f(x)\cdot g(x)=\lim_{x\to x_0}f(x)\cdot \lim_{x\to x_0}g(x)$), we get:

$$\lim_{x\to 0^+}2^{1/x}\frac{x-1}{x-2}=\infty$$

$$\lim_{x\to 0^-}2^{1/x}\frac{x-1}{x-2}=0$$

so the limit doesn't exist.

Arnaldo
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