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Expert: - 10/14/2008


Question
Hello, I recently had my first calculus exam. It was primarily a test on review material. In a study session with our TA, the question was asked how is the lim x->0 (cosx-1)/x = 0? Our TA explicitly said the process is very long and intricate and thus would NEVER be asked on the exam and rather that we should simply understand it as true and possibly apply it. However, question TWO on the exam was: "Explain why the following limit is zero: lim x->0 (cosx-1)/x". I assumed the Squeeze Theorem needed to be used but didn't really know what to bound it by. Thank you

Answer
One way to see what the limit is: Look at the Taylor's theorem for cos(x).  The Taylor's theorem states that
cos(x)=sum(k=0 to ∞)((-x)^(2k+1)/(2k+1)!.

When this is divided by x, it's sum(k=0 to ∞)((-x)^(2k)/(2k+1)! (note that power of x reduce by one.  This means that as x goes to 0, all of the terms go to 0 except for the first one.  This is because the first one is (-x)^0/1!.  Anything to the 0 is 1 and 1! is also 1.  The answer is therefore 1.

There might be another way to explain it, but that's the one I thought of right now.

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