X- 0limlog2-x-log2-xx

lim _ x→ 0 log (2+x) log (2 x) #160; #160; / x #160; This is 00 form . So by using L hospital rule lim _ x→ 0 1 / 2+x + 1 / 2 x #16 ...

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Indeterminate Forms

x→ 0limlog2+x...

Question

$lim x \to 0 \frac{log (2 + x) - log (2 - x)}{x}$

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Solution

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l o g_{2} x + l o g_{2} (\sqrt{x}) + l o g_{2} (\sqrt[4]{x}) + l o g_{2} (\sqrt[8]{x}) + l o g_{2} (\sqrt[16]{x}) + . . . . . = 4

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

x > 0

l o g_{2} x + l o g_{2} (\sqrt{x}) + l o g_{2} (\sqrt[4]{x}) + l o g_{2} (\sqrt[8]{x}) + . . . \infty = 4,

x =

f (x) = \frac{l o g (2 + x) - l o g (2 - x)}{t a n x}, f o r x \neq 0 = 1, f o r x = 0.

x = 0

f (x) = \frac{l o g \frac{1}{2} + l o g (2 + 4 x)}{x}

x \neq 0

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