Is there any way to evaluate
lim
without using L'Hopital's Rule? I was trying to use some of the standard trig limits (e.g. \lim_{x\to 0} \frac{\sin x}{x} = 1), etc. but couldn't figure it out.
Thank you.
Answer
Adding and subtracting x in nummerator we can split the expression under limit as \frac{\cos x-1} {x} +\frac{x-\sin x} {x^2} The first expression tends to 0 because (1-\cos x) /x^2\to 1/2 and the second expression also tends to 0 as shown below.
Since the second expression is an odd function it is sufficient to prove that the expression tends to 0 as x\to 0^{+}. Next note the famous inequality $$\sin x
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