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$$\displaystyle \lim_{x\to 1}\left(\frac{x^2+x+1}{2x+1}\right)^{\frac{1}{x^2-1}}$$ I know that the two-sided limit of $\frac{1}{x^2-1}$ does not exist. I don't know what to do with $\frac{x^2+x+1}{2x+1}$ to get something else than $1^\infty$

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    $\begingroup$ Did you try to rewrite this as a function of $t=x-1$? $\endgroup$
    – Did
    Commented Jan 11, 2015 at 11:28
  • $\begingroup$ No, I'll try it now. $\endgroup$
    – miko
    Commented Jan 11, 2015 at 11:30
  • $\begingroup$ possible duplicate of What is the limit of $\endgroup$
    – terrace
    Commented Jan 11, 2015 at 11:36
  • $\begingroup$ @TerrenceTown, why is it a duplicate? It's another limit. $\endgroup$
    – Alex Silva
    Commented Jan 11, 2015 at 11:37
  • $\begingroup$ @miko, If you cannot use Taylor series, you need to specify this in your question body. People take time to help you with some ideas and, only after their answers, you say that their methods cannot be used. This is not pleasant at all. $\endgroup$
    – Alex Silva
    Commented Jan 11, 2015 at 13:08

3 Answers 3

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Use : $$\lim\limits_{x \to 1}h(x)^{g(x)}=\lim\limits_{x \to 1}\mathbb{e}^{g(x)\ln{h(x)}}$$

In your case, $\lim\limits_{x \to 1}g(x)\ln{h(x)}$ is pretty easy :

Using taylor series : $$\ln h(x) = \frac{(x-1)}{3}+\frac{(x-1)^2}{18} + o((x-1)^2)$$

Then factor a bit : $$\lim\limits_{x \to 1}g(x)\ln{h(x)}=\lim\limits_{x \to 1}\frac{x+5}{18(x+1)}=\frac{1}{6}$$

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  • $\begingroup$ Could you explain, if you downvote :? $\endgroup$
    – servabat
    Commented Jan 11, 2015 at 11:49
  • $\begingroup$ I will undone my downvote if you show that $\lim \limits_{x\rightarrow 1}g(x) \text{ln}\hspace{1mm} h(x)$ is pretty easy. $\endgroup$
    – Alex Silva
    Commented Jan 11, 2015 at 11:50
  • $\begingroup$ Taylor series, and you find $\frac{1}{6}$ $\endgroup$
    – servabat
    Commented Jan 11, 2015 at 11:51
  • $\begingroup$ Ok, you should write that. "pretty easy" is meaningless. $\endgroup$
    – Alex Silva
    Commented Jan 11, 2015 at 11:51
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    $\begingroup$ I don't think OP is looking for someone doing the work for him. That's why I just wanted to give an "hint" $\endgroup$
    – servabat
    Commented Jan 11, 2015 at 11:53
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Hint

Consider $$A=\left(\frac{x^2+x+1}{2x+1}\right)^{\frac{1}{x^2-1}}$$ Taking logarithms $$\log(A)=\frac{1}{x^2-1}\log\left(\frac{x^2+x+1}{2x+1}\right)$$ Now, consider the Taylor series built at $x=1$ $$\frac{x^2+x+1}{2x+1}=1+\frac{x-1}{3}+\frac{1}{9} (x-1)^2+O\left((x-1)^3\right)$$ and use the fact that, for small $y$, $$\log(1+y)=y-\frac{y^2}{2}+O\left(y^3\right)$$ Now, make $$y=\frac{x-1}{3}+\frac{1}{9} (x-1)^2$$ in order to get the expansion of $\log\left(\frac{x^2+x+1}{2x+1}\right)$ and simplify.

I am sure that you can take from here and conclude not only what is the limit but also how it is approached.

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Let $y=x-1$. Then we have $$\require{cancel}\lim_{y\to0}\left(\frac{y^2+3y+3}{y^2-(y+1)^2+4y+4}\right)^{\frac{1}{y(y+2)}}=\lim_{y\to0}\left(\frac{y^2+3y+3}{2y+3}\right)^{1/(2y)}=\\ \lim_{y\to0}\left(\frac{\cancel{y}(y+3+3/y)}{\cancel{y}(2+3/y)}\right)^{1/(2y)},$$ which, letting $z=1/y$, yields $$\lim_{z\to\infty}\left(\frac{3z+1/z+3}{3z+2}\right)^{z/2}=\lim_{z\to\infty}\left(1+\frac{1+1/z}{3z+2}\right)^{z/2}=\lim_{z\to\infty}\left(1+\frac{1}{3z+2}\right)^{z/2}.$$ Now we could do another substitution, but it is easy to see that $2$ next to $3z$ doesn't count when $z\to\infty$. Thus we can finally conclude the limit is $$\lim_{z\to\infty} \left(1+\frac{1}{3z}\right)^{z/2}=e^{1/6}.$$

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