# how to solve this two limit tasks

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

• Hey! Your question is unreadable. Please try to edit it using MathJax. I tried but found some parts that are a bit cryptic Feb 19 '19 at 22:22
• The limit is $\sqrt{x+1} - \sqrt[4]{x^2 + x +1}x^A$??? and $x \rightarrow ??$ Feb 19 '19 at 22:22
• lim goes to + infinity ,thank you for mentioning Feb 19 '19 at 22:28

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 $$a^3-b^3=(a-b)(a^2+ab+b^2)$$,$$\forall a,b \in \mathbb{R}$$.

• 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) Feb 20 '19 at 0:15
• I used the formula I mentioned in the note. In your case you have $a=x+1$ and $b=\sqrt[3]{x^3+x^2}$. Feb 20 '19 at 14:13
• ok i understand it tnx Feb 20 '19 at 17:11

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.

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