How to find the limit of this function?

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OmarMohamedKhallaf

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[MATH]\lim\limits_{x\to 0}\frac{2x+sin(3x)}{tan(5x)}[/MATH]I evaluated the expression like this:
[MATH]\frac{2x+sin(3x)}{tan(5x)} = \frac{2x}{tan(5x)}+\frac{sin(3x)}{tan(5x)}[/MATH][MATH]\lim\limits_{x\to 0}\frac{2x+sin(3x)}{tan(5x)} = \lim\limits_{x\to 0} \frac{2x}{tan(5x)}+\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}=\frac{2}{5}+\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}[/MATH]and I'm stuck here, how should I complete?
 
Try to get 3x under the sin(3x) and get 5x above the tan(5x) and also break up tan(5x) into sin(5x)/cos(5x)
 
OmarMohamedKhallaf said:
[MATH]\lim\limits_{x\to 0}\frac{2x+sin(3x)}{tan(5x)}[/MATH]I evaluated the expression like this:
[MATH]\frac{2x+sin(3x)}{tan(5x)} = \frac{2x}{tan(5x)}+\frac{sin(3x)}{tan(5x)}[/MATH][MATH]\lim\limits_{x\to 0}\frac{2x+sin(3x)}{tan(5x)} = \lim\limits_{x\to 0} \frac{2x}{tan(5x)}+\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}=\frac{2}{5}+\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}[/MATH]and I'm stuck here, how should I complete?
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Do not waste time! Use L'Hôpital's rule.
 
Thank you, this was the missing part:
[MATH]\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}=\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}\times \frac{3x}{3x}\times \frac{5x}{5x}[/MATH]I didn't need to break [MATH]tan(5x)[/MATH]
 
OmarMohamedKhallaf said:
[MATH]\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}=\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}\times \frac{3x}{3x}\times \frac{5x}{5x}[/MATH]I didn't need to break [MATH]tan(5x)[/MATH]
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Omar, ole man, do you know L'Hôpital's rule.?
 
Honestly no, but I found an article on Wikipedia and it seems to have something related to differentiation equations which I haven't studied yet.
 
OmarMohamedKhallaf said:
Thank you, this was the missing part:
[MATH]\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}=\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}\times \frac{3x}{3x}\times \frac{5x}{5x}[/MATH]I didn't need to break [MATH]tan(5x)[/MATH]
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So what is the answer?
 
[MATH]\lim\limits_{x\to 0}\frac{2x+sin(3x)}{tan(5x)}=\lim\limits_{x\to 0}\frac{2x}{tan(5x)}+\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}[/MATH][MATH]\lim\limits_{x\to 0}\frac{2x}{tan(5x)}=\frac{2}{5}[/MATH][MATH]\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}=\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}\times\frac{3x}{3x}\times\frac{5x}{5x}=\frac{3x}{5x}=\frac{3}{5}[/MATH]There sum: [MATH]\frac{2}{5}+\frac{3}{5}=1[/MATH]
 
OmarMohamedKhallaf said:
[MATH]\lim\limits_{x\to 0}\frac{2x+sin(3x)}{tan(5x)}=\lim\limits_{x\to 0}\frac{2x}{tan(5x)}+\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}[/MATH][MATH]\lim\limits_{x\to 0}\frac{2x}{tan(5x)}=\frac{2}{5}[/MATH][MATH]\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}=\lim\limits_{x\to 0}\frac{sin(3x)}{tan(5x)}\times\frac{3x}{3x}\times\frac{5x}{5x}=\frac{3x}{5x}=\frac{3}{5}[/MATH]There sum: [MATH]\frac{2}{5}+\frac{3}{5}=1[/MATH]
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I was hoping that you can tell us how you got 3/5? It seems that you are just ignoring the 3x/3x + 5x/5x. If that is the case, then why do you need them?
 
Do you mean I'm ignoring [MATH]\frac{3x}{3x}\times\frac{5x}{5x}\ (\times \text{ not } +)[/MATH] ?
If so, then no I'm not ignoring them.
[MATH]\lim\limits_{x\to 0}\frac{tan(x)}{x} = 1\,, \lim\limits_{x\to 0}\frac{sin(x)}{x} =1[/MATH][MATH]\text{so }\frac{sin(3x)\times 3x}{tan(5x)\times 3x}=\frac{3x}{tan(5x)}[/MATH]and the same for [MATH]\frac{3x\times 5x}{tan(5x)\times 5x}=\frac{3x}{5x}[/MATH]
 
OmarMohamedKhallaf said:
Do you mean I'm ignoring [MATH]\frac{3x}{3x}\times\frac{5x}{5x}\ (\times \text{ not } +)[/MATH] ?
If so, then no I'm not ignoring them.
[MATH]\lim\limits_{x\to 0}\frac{tan(x)}{x} = 1\,, \lim\limits_{x\to 0}\frac{sin(x)}{x} =1[/MATH][MATH]\text{so }\frac{sin(3x)\times 3x}{tan(5x)\times 3x}=\frac{3x}{tan(5x)}[/MATH]and the same for [MATH]\frac{3x\times 5x}{tan(5x)\times 5x}=\frac{3x}{5x}[/MATH]
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OK.
 
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