the limit, as x -> 0, of [tan(3x) + 3x] / [sin(5x)]

[cherryontop921]

the limit as x approaches zero of:

[tan(3x) + 3x] / sin(5x)

Because when I plug 0 in for x in this equation I get 0/0, this is a clue to change or simplify the equation in some way. I tried using the "special limits", i.e. the limit as x approaches 0 of sinx/x is 1, and the limit as x approaches 0 of (1-cosx)/x is 0, but I did not get anywhere. How do I solve this problem?

 

[galactus]

Rewrite as:

\(\displaystyle \L\\\frac{\frac{sin(3x)}{cos(3x)}}{sin(5x)}+\frac{3x}{sin(5x)}\)=\(\displaystyle \L\\\frac{sin(3x)}{sin(5x)}\cdot\frac{1}{cos(3x)}+\frac{3x}{sin(5x)}\)

\(\displaystyle \L\\\frac{\frac{3sin(3x)}{3x}}{\frac{5sin(5x)}{5x}}\cdot\frac{1}{cos(3x)}+\frac{\frac{3x}{x}}{\frac{5sin(5x)}{5x}}\)

\(\displaystyle \L\\\frac{3}{5}\cdot{1}+\frac{3}{5}\)

 

[soroban]

Hello, cherryontop921!

Galactus gave an excellent explanation . . . did you follow it?

\(\displaystyle \L\lim_{x\to0}\frac{\tan 3x\,+\,3x}{\sin5x}\)

Your game plan is a good one.

We will use: \(\displaystyle \L\:\lim_{\theta\to0}\left(\frac{\sin\theta}{\theta}\right) \:=\:1\;\;\) and \(\displaystyle \L\;\;\lim_{\theta\to0}\left(\frac{\theta}{\sin\theta}\right)\:=\:1\)

And we must hammer the expression into these forms . . .


Make two fractions: \(\displaystyle \L\:\frac{\tan3x}{\sin5x}\,+\,\frac{3x}{\sin5x}\)


The first fraction is: \(\displaystyle \L\:\frac{\sin3x}{1}\,\cdot\,\frac{1}{\cos3x}\,\cdot\,\frac{1}{\sin5x}\)

Multiply the first by \(\displaystyle \frac{3x}{3x}\), the third by \(\displaystyle \frac{5x}{5x}:\;\;\L\frac{3x}{3x}\,\cdot\,\frac{\sin3x}{1}\,\cdot\,\frac{1}{\cos3x}\,\cdot\,\frac{5x}{5x}\,\cdot\,\frac{1}{\sin5x}\)

. . which rearranges to: \(\displaystyle \L\:\frac{3}{5}\,\cdot\,\frac{\sin3x}{3x}\,\cdot\,\frac{1}{\cos3x}\,\cdot\,\frac{5x}{\sin5x}\)


Multiply the second fraction by \(\displaystyle \frac{5x}{5x}:\;\;\L\frac{5x}{5x}\,\cdot\,\frac{3x}{\sin5x}\)

. . which rearranges to: \(\displaystyle \L\:\frac{3}{5}\,\cdot\,\frac{5x}{\sin5x}\)


So we have: \(\displaystyle \L\:\lim_{x\to0}\left[\frac{3}{5}\,\cdot\,\frac{\sin3x}{3x}\,\cdot\,\frac{1}{\cos3x}\,\cdot\,\frac{5x}{\sin5x}\:+\:\frac{3}{5}\,\cdot\,\frac{5x}{\sin5x}\right]\)

. . . . . . . . \(\displaystyle \L=\;\frac{3}{5}\,\cdot\,1\,\cdot\,\frac{1}{1}\,\cdot\,1 \;+\;\frac{3}{5}\,\cdot\,1\)

. . . . . . . . \(\displaystyle \L= \;\frac{3}{5}\,+\,\frac{3}{5} \;=\;\frac{6}{5}\)

 

[cherryontop921]

Thank you all so much! I follow what you're saying.