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

I just started limits as part of my calculus class, and I have a simple limit to evaluate, $\mathop{\lim}\limits_{x \to 1}~f(x)$ . I found the limit to be 2 using the $f(x)=\mathop{\lim}\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$ method. However, upon checking my result in Mathematica (as well as Wolfram|Alpha), the result is shown as being 1. Is this a problem in Mathematica, or am I doing something wrong?

EDIT: $f(x)=x^2$

share|cite|improve this question
what is the problem? – Yimin Feb 16 '13 at 21:31
What is $f$? In one place, you seem to be computing the limit of $f$ as $x$ goes to $1$, but in another you are computing the derivative. Could you provide more detail? – Potato Feb 16 '13 at 21:31
Without telling us what $f$ and $x$ is, how do you expect anyone to be able to answer this question? – mrf Feb 16 '13 at 21:33
...relative likelihood of novice making mistake versus wildly popular software with 30 year history making mistake on trivial problem? – Jonathan Feb 16 '13 at 21:40
Maybe you want $\lim_{x\to 1} f'(x)$? – Pedro Tamaroff Feb 16 '13 at 21:41
up vote 2 down vote accepted

Well, for the function $f(x)=x^2$, we have both $$\lim_{x\to 1}\,f(x)=1\qquad\qquad \lim_{h\to 0}\,\frac{f(1+h)-f(1)}{h}=2.$$ Why would you expect them to give you the same answer?

share|cite|improve this answer
No problem. I'm wondering now: maybe he wants $\lim\limits_{x\to 1} f'(x)$? – Pedro Tamaroff Feb 16 '13 at 21:41
it should be $f(1+h)-f(1)$ – i.a.m Feb 16 '13 at 21:42
That's already been corrected, i.a.m. – Zev Chonoles Feb 16 '13 at 21:44

f is defined at 1. Therefore $f(1)=1^2=1$

Now $f(x)\not =f'(x)=\mathop{\lim}\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$ and $f'(1)=2$

share|cite|improve this answer

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.