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Without using L'Hopital's rule, create the end value: $$\lim_{x\to 0}\frac{x-\sin x}{x-\tan x}$$

5xum
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Afif bakr
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  • https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Jul 30 '18 at 08:50
  • Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Jul 30 '18 at 09:01
  • Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Jul 30 '18 at 09:01
  • Oh, and another thing. Please read this before posting: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference to better format your questions in future. I fixed the formatting of this first question, but you should not expect others to do this for you! – 5xum Jul 30 '18 at 09:01
  • If you search with Approach0, you can find some posts about this limit. For example,$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital. – Martin Sleziak Aug 30 '18 at 14:04

1 Answers1

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$$ \frac{x-\sin\left(x\right)}{x-\tan\left(x\right)}\underset{0}{=}\frac{x-x+x^3/6+o\left(x^3\right)}{x-x-x^3/3+o\left(x^3\right)}\underset{x\rightarrow 0}{\rightarrow}-\frac{3}{6}=-\frac{1}{2} $$

Atmos
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