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Find $$\lim _{x\to 0}\frac{1-\cos x}{x+x^2}$$

Give your answer as an exact number.

So I did this and got .5 as my answer. What does it mean by "exact number"? .5 was not a correct answer so I am stuck, help? thanks!

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Well what did you do? After the first step you should have $\lim\limits_{x\to 0} \frac{\sin(x)}{1+2x}$, can you tell what this limit is? –  Stefan Mar 6 '13 at 21:37
"Exact number" refers to a number that is not rounded. For example, if the answer is $\pi$ (it's not), don't input $3.141592654$, as that's not exact enough. –  anorton Mar 6 '13 at 21:39
That limit would be 0 wouldn't it? because it isn't 0/0 though, would 0 be an appropriate answer or would I need to apply L'Hopitals rule again? –  user59714 Mar 6 '13 at 21:39
The answer happens not to be exactly right, it should be $0$. After one application of L'Hospital's Rule, you no longer have an indeterminate form. So in this case, applying L'H. a second time is wrong. –  André Nicolas Mar 6 '13 at 21:41
You can apply the rule if you have an expression like $\frac 0 0$ or $\frac \infty \infty$. And yes, if you have a limit, you are finished. –  Stefan Mar 6 '13 at 21:41

1 Answer 1

You know by L'Hopital: $$ \lim_{x\to 0}\frac{1-\cos(x)}{x+x^2} = \lim_{x\to 0}\frac{\sin(x)}{1+2x} $$ And what is the limit of $\sin(x)$ for $x \to 0$ ?

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the limit would be 0, thank you :) –  user59714 Mar 6 '13 at 21:44

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