Question #041b9

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Question #041b9

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Answer 1 · 2017-05-14T15:34:46

1

Explanation:

$\frac{2^{\frac{1}{x}}}{1 + 2^{\frac{1}{x}}} = \frac{2^{u}}{1 + 2^{u}}$ $2^(1/x)/(1+2^(1/x)) = 2^u/(1+2^u)$ where $u$ $u$ approaches to $+ \infty$ $+oo$

$\frac{2^{u}}{1 + 2^{u}} = \frac{1}{\frac{1}{2^{u}} + 1}$ $2^u/(1+2^u) = 1/(1/(2^u)+1)$

$u \to + \infty \Leftrightarrow 2^{u} \to + \infty$ $u -> +oo <=> 2^u -> +oo$

then the limit will be $\frac{1}{0 + 1} = 1$ $1/(0+1) = 1$

Answer 2 · 2017-05-14T15:42:41

$1 .$ $1.$

Explanation:

As $x \to 0 +, \frac{1}{x} \to + \infty .$ $x to 0+, 1/x to +oo.$

Recall that, $as x \to \infty +, a^{x} \to \infty +, where a > 1 .$ $"as "x to oo+, a^x to oo+," where "a gt 1.$

$:. 2^(1/x) to oo+," & so, "(1+2^(1/x)) to oo+," giving, "$

$lim_(x to 0+) 1/(1+2^(1/x))=0................(ast).$

$"Now, "2^(1/x)/(1+2^(1/x))={(1+2^(1/x))-1}/(1+2^(1/x))$

$=(1+2^(1/x))/(1+2^(1/x))-1/(1+2^(1/x))$

$=1-1/(1+2^(1/x)).$

$:.," by "(ast)," The Reqd. Lim.="1-0=1.$

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Question #041b9

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