Prof says answer is 5 to the following: lim as x goes to 0 of (sin 5x)/x. is he right and how does it work?

hi big nose

let u = 5x

as x ->0 , u -> 5*0 = 0 , and u = 5x so u/5 = x now we have lim u-> 0 sin u / ( u/5) = lim 5 * sin u/u

we know that lim sin x / x as x goes to zero is 1, this is a known fact or theorem. you can prove it seperately using squeeze theorem or by geometry

so lim 5 * sin u/u = 5 * lim sin u/u = 5 * 1

\[\lim_{x \rightarrow 0} \sin(5x)/x\] produces the indeterminate form 0/0, so you can use L'Hopital's rule that: \[\lim_{x \rightarrow a}f(x)/g(x) = \lim_{x \rightarrow a} f \prime (x) / g \prime (x)\] (keep in mind that you can only use this if the original limit produces 0/0 or infinity/infinity!). So, applying the rule gives us: \[\lim_{x \rightarrow 0}5\cos(x)\] cos(0) is one, so the answer is 5.

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