limits: why does the limit of sin(5x)/x equal 5?

[mathishard124]

why does the lim as x goes to 0 of sin(5x)/x equal 5?


i know that the limit of sin(x)/x = 1, but I don't see how this is the constants limit law, becuase sin (5x) is not the same as 5sin(x).

 

[skeeter]

lim{x -> 0} sin(5x)/x =

lim{x->0} 5*sin(5x)/(5x) =

5*lim{x->0} sin(5x)/(5x)

as x->0, 5x->0 ...

so, 5*lim{x->0} sin(5x)/(5x) = 5(1) = 5

 

[soroban]

Hello, mathishard124!

why does the lim as x goes to 0 of sin(5x)/x equal 5?

You already know that: \(\displaystyle \L\,\lim_{\theta\to0}\frac{\sin\theta}{\theta}\:=\:1\)

Note that the three \(\displaystyle \theta\)'s are identical.

So: \(\displaystyle \L\,\lim_{5q\to0}\frac{\sin(5q)}{5q}\:=\:1\)

and: \(\displaystyle \L\,\lim_{7n\to0}\frac{\sin(7n)}{7n} \:=\:1\)


However, in \(\displaystyle \L\,\lim_{x\to0}\frac{\sin(5x)}{x}\), the three variables are not identical.

But we can make them identical . . .


Multiply the fraction by \(\displaystyle \frac{5}{5}:\L\;\lim_{x\to0}\,\left[\frac{5}{5}\cdot\frac{\sin(5x)}{x}\right] \;= \;\lim_{x\to0}\left[5\cdot\frac{\sin(5x)}{5x}\right]\)


We know that if \(\displaystyle x\to0\), then \(\displaystyle 5x\to0\)

So we have: \(\displaystyle \L\,\lim_{5x\to0}\left[5\cdot\frac{\sin(5x)}{5x}\right] \;= \;5\cdot\underbrace{\left[\lim_{5x\to0}\frac{\sin(5x)}{5x}\right]} \;=\;5\cdot1\;=\;5\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . This is \(\displaystyle 1\)