#lim sin(2x)/x

13 messages · Page 1 of 1 (latest)

stable crescent
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where is the flaw in the following reasoning:
$$lim_{x\to0}\frac{sin(2x)}{x}=lim_{x\to0}\frac{sin(2x)}{2x}\cdot x=lim_{2x\to0}\frac{sin(2x)}{2x} \cdot lim_{x\to0}x = 1\cdot 0$$

gentle jayBOT
stable crescent
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it cut off at the end

#

$$lim{x\to0}\frac{sin(2x)}{x}=lim{x\to0}\frac{sin(2x)}{2x}\cdot x=lim{2x\to0}\frac{sin(2x)}{2x} \cdot lim{x\to0}x$$

fierce ingotBOT
stable crescent
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$$= 1\cdot 0$$

fierce ingotBOT
frail thistle
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From the first step to the second, you multiplied the numerator by x, but the denominator by 2

stable crescent
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im dumb

frail thistle
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But you asked a specific question and showed your work, which is something I'm always trying to teach everyone here.

stable crescent
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yea i know it's just the proofs part at the start of this calculuscourse made me forget basic algebra somehow

stark juniper
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Dont call yourself dumb for making a mistake