# What Is the Limit of (x - tan(x)) / x³ as x Approaches 0?

• navneet9431
In summary, you cannot calculate with undetermined expressions like ##\frac{1}{x^2}## as if they were numbers.
navneet9431
Gold Member
<Moderator's note: Moved from a technical forum and thus no template.>
$$\lim_{x\rightarrow 0} (x-tanx)/x^3$$
I solve it like this,
$$\lim_{x\rightarrow 0}1/x^2 - tanx/x^3=\lim_{x\rightarrow 0}1/x^2 - tanx/x*1/x^2$$
Now using the property $$\lim_{x\rightarrow 0}tanx/x=1$$,we have ;
$$\lim_{x\rightarrow 0}1/x^2 - 1/x^2=0$$

Last edited by a moderator:
navneet9431 said:
$$\lim_{x\rightarrow 0} (x-tanx)/x^3$$
I solve it like this,
$$\lim_{x\rightarrow 0}1/x^2 - tanx/x^3=\lim_{x\rightarrow 0}1/x^2 - tanx/x*1/x^2$$
Now using the property $$\lim_{x\rightarrow 0}tanx/x=1$$,we have ;
$$\lim_{x\rightarrow 0}1/x^2 - 1/x^2=0$$

No, it is wrong. You cannot calculate with undetermined expressions like ##\frac{1}{x^2}## as if they were numbers.
The shortest way is possibly a Taylor expansion of the function. Which theorems are you supposed to use?

No you are wrong, you have to be carefull when you do algebra with limits, you cannot go back and forth the way you do

Instead use L'Hopital once and you ll calculate the correct value which is ##\frac{-1}{3}##

navneet9431 said:
$$\lim_{x\rightarrow 0}tanx/x=1$$,we have ;
$$\lim_{x\rightarrow 0}1/x^2 - 1/x^2=0$$
To be a bit more exact regarding your error, you cannot in general turn lim(a+bc) into lim(a+c lim(b)).
The reason is that in taking the inner limit you may discard a second order term, but after multiplying what remains by c the result may then cancel with a, so the term discarded was important after all.
You may be able to rescue your method by keeping more terms and using the O() notation.

haruspex said:
you may discard a second order term
About what second order term are you talking?
Note: English is my second language.
I will be thankful for any help!

navneet9431 said:
About what second order term are you talking?
Note: English is my second language.
I will be thankful for any help!
More general than saying
$$\lim_{x\rightarrow 0}tanx/x=1$$,
you can write tan(x)/x=1+x2/3+O(x4).
This way, after some cancellation, you have a nonzero leading term.

When all the limits involved are finite, then the algebra with limits is pretty much the same as the algebra with real numbers.

However when some limits are infinite then the algebra of limits might break down, and in this example the limits ##lim\frac{1}{x^2}## and ##\lim-\frac{1}{x^2}## which one has limit ##+\infty## and the other ##-\infty## (assuming we take limits ##x\to 0+##) and they cannot be recombined as the last step at the OP does, the rule for the sum of limits breaks down when one limit is ##+\infty## and the other is ##-\infty##.

## 2. Can I trust that my answer is correct?

Yes, you can trust that your answer is correct if it has been verified by a reliable source or if you have used a reliable method to arrive at your answer.

## 3. What if my answer is different from the expected answer?

If your answer is different from the expected answer, it is possible that you made a mistake or used a different approach to solving the problem. It is important to double check your work and make sure you understand the problem before concluding that your answer is incorrect.

## 4. Why is it important to have a correct answer?

A correct answer is important because it ensures that the information or solution being presented is accurate and can be relied upon. It also helps to build trust and credibility in the scientific community.

## 5. What if my answer is partially correct?

If your answer is partially correct, it means that you have some understanding of the problem or concept, but may have made a mistake or missed some key information. It is important to review your work and identify where you may have gone wrong in order to fully understand the concept and arrive at the correct answer.

• Calculus and Beyond Homework Help
Replies
8
Views
787
• Calculus and Beyond Homework Help
Replies
20
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
175
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
893
• Calculus and Beyond Homework Help
Replies
26
Views
2K
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K