Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – apnorton 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

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 $x \to 0$ ?

share|cite|improve this answer
the limit would be 0, thank you :) – user59714 Mar 6 '13 at 21:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.