I've been working on this problem for a long time now and I haven't been able how to figure it out. It is: lim sin x/x+tan x *as x approaches 0* The answer is 1/2 but i dont understand how to get there
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Consider $\displaystyle \lim_{x\to 0 }\frac{\sin x }{x} = 0$
Originally Posted by pickslides Consider $\displaystyle \lim_{x\to 0 }\frac{\sin x }{x} = 0$ YOU mean 1 not 0
Originally Posted by mhuggin2 I've been working on this problem for a long time now and I haven't been able how to figure it out. It is: lim sin x/x+tan x *as x approaches 0* The answer is 1/2 but i dont understand how to get there Divide the numerator and denominator by x $\displaystyle \lim_{x\to0}\frac{\sin{x}}{x+\tan{x}}=\lim_{x\to0} \frac{\dfrac{\sin{x}}{x}}{1+\dfrac{1}{\cos{x}}\cdo t\dfrac{\sin{x}}{x}}=\frac{1}{1+\dfrac{1}{1}\cdot1 }=\frac{1}{2}.$
Originally Posted by autumn YOU mean 1 not 0 yep, silly typo.
OMG thank you SO much. It all makes so much sense now
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