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Hello i stumbled across this two limits task and i cant find an answer to them:

  1. Find the limit depending on the parameter $A$ $$\lim \limits_{x\to\infty}\left(\left(\sqrt{x+1} - \sqrt[4]{x^2 + x + 1} \right) \cdot x^A\right)$$

I tried by multipliying with $$\frac{\sqrt{x+1} + \sqrt[4]{x^2 + x + 1}}{\sqrt{x+1} + \sqrt[4]{x^2 + x + 1}}$$ which equals 1 and then $$\frac{x+1 + \sqrt{x^2 + x + 1}}{x+1 + \sqrt{x^2 + x + 1}}$$ so i can get rid off roots in denominators but i got tangled up,

  1. $$\lim\limits_{x\to\infty}\left((x+1) - \sqrt[3]{x^3 + x^2} \right)$$

lim in both tasks goes to +infinity

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  • $\begingroup$ Hey! Your question is unreadable. Please try to edit it using MathJax. I tried but found some parts that are a bit cryptic $\endgroup$ Feb 19 '19 at 22:22
  • $\begingroup$ The limit is $\sqrt{x+1} - \sqrt[4]{x^2 + x +1}x^A$??? and $x \rightarrow ??$ $\endgroup$
    – JoseSquare
    Feb 19 '19 at 22:22
  • $\begingroup$ lim goes to + infinity ,thank you for mentioning $\endgroup$
    – Petar
    Feb 19 '19 at 22:28
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For 2 we have $\lim\limits_{x\to\infty}\left((x+1) - \sqrt[3]{x^3 + x^2} \right)=\lim\limits_{x\to\infty}\frac{2x^2+3x+1}{(x+1)^2+(x+1)\sqrt[3]{x^3 + x^2} +\sqrt[3]{(x^3 + x^2)^2}}$.
Now you can factor $x^2$ both in the numerator and the denominator and see that the limit is $\frac{2}{3}$.The first limit can be done similarly.
Note: What I used was that $a^3-b^3=(a-b)(a^2+ab+b^2)$,$\forall a,b \in \mathbb{R}$.

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    $\begingroup$ Thank you but how did you come from (x+1) - 3th root(xpow3 + xpow2) to lim 2xpow2+3x + 1 /(x+1)pow2 + (x+1) 3th root(xpow3+xpow2) + 3th root((xpow3 + xpow2)pow2) $\endgroup$
    – Petar
    Feb 20 '19 at 0:15
  • $\begingroup$ I used the formula I mentioned in the note. In your case you have $a=x+1$ and $b=\sqrt[3]{x^3+x^2}$. $\endgroup$
    – Alexdanut
    Feb 20 '19 at 14:13
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    $\begingroup$ ok i understand it tnx $\endgroup$
    – Petar
    Feb 20 '19 at 17:11
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You should be aware of Taylor expansion. Substitute $x=1/t$, so $t\to0^+$ and the limit can be rewritten as $$ \lim_{t\to0^+}\frac{\sqrt{1+t}-\sqrt[4]{1+t+t^2}}{t^{A+1/2}} $$ The numerator can be rewritten as $$ 1+\frac{1}{2}t-1-\frac{1}{4}{t}+o(t) $$ So the limit is $1/2$ when $A+1/2=1$. What if $A>-1/2$ or $A<-1/2$?

The second limit can be computed similarly.

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  • $\begingroup$ thank you but what does the o(t) represents? $\endgroup$
    – Petar
    Feb 20 '19 at 0:07
  • $\begingroup$ @Petar $o(t)$ is an unnamed function with the property that $\lim_{t\to0}\frac{o(t)}{t}=0$. $\endgroup$
    – egreg
    Feb 20 '19 at 0:26

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