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limit as x approaches 0 of (sin 5x)/7x?
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I've considered multiply the numerator by 5x/5x so that sin x will become 1 as x approaches zero...but doesn't the problem then become 5x/7x and they both go to zero?
\[\lim_{x \rightarrow 0}\frac{\sin(5x)}{5x} =1\] right? so how do we get this in our problem? \[\frac{5}{7}\lim_{x \rightarrow 0}\frac{\sin(5x)}{5x}\]
wouldn't there be no limit? *confuzzled* O.o
\[\frac{5}{7}(1)=\frac{5}{7}\]
Sorry, is there an intermediate step missing?
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You can verify that myininaya is right very quickly if you know L'hospital's rule.
all i did was multiply 5/5 @ josh
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