Evaluate the limit-limx - 08x-4x-2x-1x2

Given: lim_x → 08^x 4^x 2^x+1x^2 =lim_x → 02^3 x 2^2 x 2^x+1x^2 =lim_x → 02^2 x(2^x 1) 1(2^x 1)x^2 =lim_x → 0(2^2 x 1)(2^x 1)x^2 =lim_x → 0 2 ×(2^2 x 1)2 x× ...

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Standard XII

Mathematics

Summation Using Sigma

Evaluate the ...

Question

Evaluate the limit:
$lim x \to 0 \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}$

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Solution

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

$= lim x \to 0 \frac{2^{3 x} - 2^{2 x} - 2^{x} + 1}{x^{2}}$

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

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

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

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

$= 2 log 2 \times log 2$
$[lim x \to 0 (\frac{a^{x} - 1}{x}) = log a]$

$= (log 2^{2}) (log 2)$
$= (log 4) (log 2)$

Therefore,
$lim x \to 0 \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}} = (log 4) (log 2)$

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Summation Using Sigma

Standard XII Mathematics

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Evaluate the limit: limx→08x−4x−2x+1x2

Evaluate the limit:
$lim x \to 0 \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}$