Lim x - 02x-1-1-x1-2-1-

The correct option is B log 4lim_x → 02^x 1/(1+x)^1/2 1= lim_x → 02^xlog 2/1/2(1+x)^ 1/2 {∵ lim_x → af(x)/g(x)= lim_x → af'(x)/g'(x)} = 2 log 2 = log 4 ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Limits

lim x → 02x-1...

Question

${lim}_{x \to 0} \frac{2^{x} - 1}{(1 + x)^{\frac{1}{2}} - 1} =$

log 2

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log 4

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log

\sqrt{2}

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None of these

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Solution

The correct option is B log 4
${lim}_{x \to 0} \frac{2^{x} - 1}{(1 + x)^{\frac{1}{2}} - 1} = {lim}_{x \to 0} \frac{2^{x} l o g 2}{\frac{1}{2} (1 + x)^{- \frac{1}{2}}}$
${∵ l i m_{x \to a} \frac{f (x)}{g (x)} = l i m_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}}$
= 2 log 2 = log 4

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Similar questions

lim x \to 0 \frac{2^{x} - 1}{(1 + x)^{1 / 2} - 1} =

Q. Evaluate :

lim x \to 0 \frac{2^{x} - 1}{{(1 + x)}^{1 / 2} - 1}

Q. Evaluate

lim x \to 0 \frac{2^{x} - 1}{(1 + x)^{\frac{1}{2}} - 1}

Q. Use the formula

lim x \to 0 \frac{a^{x} - 1}{x} = log a

to find

lim x \to 0 \frac{2^{x} - 1}{{(1 + x)}^{1 / 2} - 1}

which is

= a log a

. Find

a

{lim}_{x \to 0} \frac{2^{x} - 1}{\sqrt{1 + x} - 1}

equals

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limx→02x−1(1+x)12−1=

The correct option is B log 4limx→02x−1(1+x)12−1=limx→02xlog 212(1+x)−12 {∵ limx→af(x)g(x)=limx→af′(x)g′(x)} = 2 log 2 = log 4

${lim}_{x \to 0} \frac{2^{x} - 1}{(1 + x)^{\frac{1}{2}} - 1} =$

The correct option is B log 4
${lim}_{x \to 0} \frac{2^{x} - 1}{(1 + x)^{\frac{1}{2}} - 1} = {lim}_{x \to 0} \frac{2^{x} l o g 2}{\frac{1}{2} (1 + x)^{- \frac{1}{2}}}$
${∵ l i m_{x \to a} \frac{f (x)}{g (x)} = l i m_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}}$
= 2 log 2 = log 4