lim[x->0] x^2-2x/sin3x; lim[x->pi/4] 1-cosx/x

[CarbonLTD]

1. Lim as X approaches 0, x^2-2x/ sin3x .

2. Lim as X approaches PI/4, 1-cosx/x
I know 1-cox/x=0, but is PI/4 in the domain ?

Thanks in advance !

Jim

 

[galactus]

For #1, divide by x:

\(\displaystyle \L\\\lim_{x\to\0}\frac{\frac{x^{2}}{x}-\frac{2x}{x}}{\frac{sin(3x)}{x}}=\lim_{x\to\0}\frac{x-2}{\frac{sin(3x)}{x}}\)

As you know, \(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{x}=1\), therefore, \(\displaystyle \L\\\lim_{x\to\0}\frac{sin(3x)}{x}=3\)

Then you have:

\(\displaystyle \L\\\lim_{x\to\0}\frac{x-2}{3}=\frac{-2}{3}\)


#2: This one doesn't take much manipulation.

\(\displaystyle \L\\\frac{1-cos(\frac{\pi}{4})}{\frac{\pi}{4}}=\frac{4-2\sqrt{2}}{\pi}\)

 

[CarbonLTD]

For #1, divide by x:

\(\displaystyle \L\\\lim_{x\to\0}\frac{\frac{x^{2}}{x}-\frac{2x}{x}}{\frac{sin(3x)}{x}}=\lim_{x\to\0}\frac{x-2}{\frac{sin(3x)}{x}}\)

As you know, \(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{x}=1\), therefore, \(\displaystyle \L\\\lim_{x\to\0}\frac{sin(3x)}{x}=3\)

Then you have:

\(\displaystyle \L\\\lim_{x\to\0}\frac{x-2}{3}=\frac{-2}{3}\)


#2: This one doesn't take much manipulation.

\(\displaystyle \L\\\frac{1-cos(\frac{\pi}{4})}{\frac{\pi}{4}}=\frac{4-2\sqrt{2}}{\pi}\)

Thanks for the quick reply ! Can you explain how you got #2 a little more? Im still alittle confused for you got 2 square roots of 2. Thanks again!