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What is the limit as x approaches zero of sin(ax)/bx?
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HI!!
make a guess i bet you will be right
I was thinking 1
@misty1212
oh no if it was \[\frac{\sin(x)}{x}\] then it would be 1
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but it is \[\frac{\sin(ax)}{bx}\]
you want a method? it is kind of silly
I am using the l'hopitals rule
oh hell no
like killing a fly with a 45
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\[\frac{\sin(ax)}{bx}=\frac{\sin(ax)}{ax}\times \frac{a}{b}\]
I dont understand
that is just some silly algebra to show you that it is \(\frac{a}{b}\) since \[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]
i will bet to a/b
so you get \[1\times \frac{a}{b}\]
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well seems you already solved this hehe
you can use l'hopital too but it is dumb \[\frac{a\cos(ax)}{b}\]
so then it would be a/b?
\[\color\magenta\heartsuit\]
okay thank you!
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yw
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