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$\lim_{x\to -1} \ \, \frac{1}{{1+x}} \left( \frac{1}{{x+5}}+ \frac{1}{{3x-1}} \right) $

As, the limit is not of the form$\ \frac{0}{{0}} $ so, put $\ x $ as $\ -1 $ we get Answer $\ 0$ . What is wrong in this?

joey rohan
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  • How is it not of the form $\frac{1}{0} \times 0$? – Calvin Lin Jul 04 '13 at 15:53
  • @CalvinLin yeah, edited' – joey rohan Jul 04 '13 at 15:55
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    The closing voter made me curious: "but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise." Really? Or is this the new whatever-does-not-fit-closing-reason? – Julien Jul 04 '13 at 15:57
  • @joeyrohan Why would you plug in 0 for $x$ while taking a limit as $x\rightarrow-1$? – Nick Peterson Jul 04 '13 at 16:11
  • @nrpeterson: I am hoping that that was a typo and they intended to "put $x$ as $-1$". – robjohn Jul 04 '13 at 16:13
  • I think the closing voter was (rather naively) closing on the basis of the question in the title. I do think the OP should change the title to something referencing the question though. – Dan Rust Jul 04 '13 at 16:17
  • @nrpeterson I made a mention of that originally, and OP replied with "edited" and so I removed the comment. However, it has not been edited in. – Calvin Lin Jul 04 '13 at 16:17
  • @DanielRust I am fairly confident the title question can be answered on the basis of precise facts. – Julien Jul 04 '13 at 16:22
  • @julien Either I'm missremembering the title of the question or it was edited to include the words 'this way'. – Dan Rust Jul 04 '13 at 16:25

2 Answers2

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If you write the limit as $$ \frac{\frac1{x+5}+\frac1{3x-1}}{x+1} $$ it becomes of the form $\frac00$. From here try combining the fractions in the numerator: $$ \frac{\frac{4x+4}{(x+5)(3x-1)}}{x+1} $$ now you can cancel the pieces that go to $0$ and try to plug in $x=-1$.

robjohn
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$\lim\limits_{x\to -1}\frac{1}{x+1}.\frac{4x+4}{(x+5)(3x-1)}=\lim\limits_{x\to -1}\frac{4}{(x+5)(3x-1)}=\frac{-1}{4}$

user5402
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