finding the limit
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No, sorry to say, it is not 2/3.
Use the identity sin(2x)=2sin(x)cos(x)
\(\displaystyle \lim_{x\to 0}\frac{sin(2x)}{3cos(x)}=\lim_{x\to 0}\frac{2sin(x)cos(x)}{3cos(x)}=\lim_{x\to 0}\frac{2sin(x)}{3}\)
Now, see the limit?. What is sin(0)?.
Also, your method was fine except \(\displaystyle \lim_{x\to 0}\frac{x}{cos(x)}=0, \;\ \text{not 2}\)
That would give you 1/3(1)(0)=0, as required.
Use the identity sin(2x)=2sin(x)cos(x)
\(\displaystyle \lim_{x\to 0}\frac{sin(2x)}{3cos(x)}=\lim_{x\to 0}\frac{2sin(x)cos(x)}{3cos(x)}=\lim_{x\to 0}\frac{2sin(x)}{3}\)
Now, see the limit?. What is sin(0)?.
Also, your method was fine except \(\displaystyle \lim_{x\to 0}\frac{x}{cos(x)}=0, \;\ \text{not 2}\)
That would give you 1/3(1)(0)=0, as required.