Lim x - 08x-4x-2x-1-x2

Lim_x→ 08^x 4^x 2^x+1/x^2 =lim_x→ 04^x(2^x 1) (2^x 1)/x^2 =lim_x→ 0(2^x 1)(4^x 1)/x^2 =lim_x→ 0 (2^x 1/x )×lim_x→ 0 (4^x 1/x ) =log 2.log 4=log 2(log2^2) =log 2 ...

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Byju's Answer

Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

lim x → 08x-4...

Question

${lim}_{x \to 0} \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}$

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Solution

${lim}_{x \to 0} \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}$

$= {lim}_{x \to 0} \frac{4^{x} (2^{x} - 1) - (2^{x} - 1)}{x^{2}}$

$= {lim}_{x \to 0} \frac{(2^{x} - 1) (4^{x} - 1)}{x^{2}}$

$= {lim}_{x \to 0} (\frac{2^{x} - 1}{x}) \times {lim}_{x \to 0} (\frac{4^{x} - 1}{x})$

$= l o g 2. l o g 4 = l o g 2 (l o g 2^{2})$

$= l o g 2 (2 l o g 2) = 2 (l o g 2)^{2}$

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Similar questions

Q. Evaluate:

{lim}_{x \to 0} \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}

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limx→08x−4x−2x+1x2

limx→08x−4x−2x+1x2 =limx→04x(2x−1)−(2x−1)x2 =limx→0(2x−1)(4x−1)x2 =limx→0(2x−1x)×limx→0(4x−1x) =log 2.log 4=log 2(log22) =log 2(2 log2)=2(log 2)2

${lim}_{x \to 0} \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}$