Why is the limit as x approaches 0 of sin(5x)/x = 5?.

lim          sin (5θ) / θ                        multiply top / bottom by 5

θ → 0

lim           5  * sin (5θ) /(5θ )     =                 

θ → 0

5 * lim      sin (5θ) / (5θ)     =    (  remember........   lim      sin x / x   = 1  )             

    θ → 0                                                                             x →  0

5  *   [1 ]   =

5

I do not know how to solve it mathematically, I hope someone does and we can both learn.

I DO know at very small values , as x approaches zero, sin(x) ~~ x     so the numerator becomes 5x as we approach zero    5x/x = 5     But I do not think that is a very good proof of your question.......

  You can GRAPH  sin(5x) / 5   and see it is true too.....still not a great proof.....

Anyone?

\(\displaystyle\lim_{\theta\rightarrow 0}\dfrac{\sin5\theta}{\theta}\\ =\displaystyle\lim_{\theta\rightarrow 0}5\cdot \dfrac{\sin 5\theta}{5\theta}\\ = 5\cdot \displaystyle\lim_{\theta\rightarrow 0} \dfrac{\sin 5\theta}{5\theta}\\ =5\cdot 1\\ = 5 \\ \boxed{\cdot\;\cdot\\ \smile}\)

This is just the visualization of CPhill's answer......

\(\lim_{\theta \rightarrow 0} \frac{sin(5\theta)}{\theta} = \frac{0}{0}\)

When dealing with limits, 0/0 basically just means that you have to find a different way of combatting the problem. CPhill's answer works under the assumption that you know \(\lim_{x\rightarrow 0}\frac{sin(x)}{x}=1\), but there is still a way to find the answer without having that memorized.

L'Hospital's Rule states that if the limit of a fraction is 0/0, then it will have the same limit as the derivative of the numerator over the derivative of the denominator.

\(\lim_{\theta\rightarrow 0} \frac{sin(5\theta)}{\theta} = \lim_{\theta \rightarrow 0} \frac{\frac{d}{d\theta}sin(5\theta)}{\frac{d}{d\theta}\theta}=\lim_{\theta\rightarrow 0} \frac{5cos(5\theta)}{1}=5cos(0)=5\)

This question is testing your knowledge of L'Hopital's Theorem.  The theorem just says that if you have a quotient of two functions,say f(x) /g(x)   where both functions have limits of zero for a particular value of x, then f(x)/ g(x)  = f'(x)/g'(x)    ie    just differentiate top and bottom.  I am using x instead of theta as I can't use the maths functions on this site for some reason.

so d/dx   sin(5x)  =  5 cos (5x)   and d/dx    x   =1

the quotient now is

5 cos(5x)   /1           let x tend to zero,    5cos(0)  = 5   and limit is   5/1  = 5.