# evaluate the limit when x goes to infinity

Evaluate $$\displaystyle{\lim_{x \to \infty} (x^3+6x^2+1)^{\frac13}-(x^2}+x+1)^{\frac12}$$

I tried writing it as $$\lim_{x \to \infty}\frac{x^3+6x^2+1}{(x^3+6x^2+1)^{\frac23}}-\frac{x^2+x+1}{(x^2+x+1)^{\frac12}}$$ but I do not know how to continue from this point.

I really appreciate your time and effort. Thank you very much!

• Hint: rewrite as$$\lim_{x\to\infty}((x^3+6x^2+1)^{1/3}-x)-\lim_{x\to\infty}((x^2+x+1)^{1/2}-x)$$then use$$a-b=\frac{a^n-b^n}{\sum_{k=0}^{n-1}a^kb^{n-1-k}}$$to evaluate each limit.
– J.G.
Nov 11, 2023 at 9:47
• Thank you very much! Nov 11, 2023 at 9:49

Let $$t=\frac{1}{x}$$. Then

$$(x^3+6x^2+1)^{\frac13}-(x^2+x+1)^{\frac12} = \frac{(t^3+6t+1)^{\frac13}-(t^2+t+1)^{\frac12}}{t} = \frac{f(t)-f(0)}{t-0}$$

where $$f(t)=(t^3+6t+1)^{\frac13}-(t^2+t+1)^{\frac12}$$. Therefore, if it exists, the limit you are looking for is equal to $$\lim_{t\to 0^+}\frac{f(t)-f(0)}{t-0}$$. Since $$f$$ is differentiable at $$0$$, the limit exists and is equal to $$f'(0)$$.

Using Taylor series, $$f(t)=0+\frac{3}{2}t+...$$, so $$f'(0)=\frac{3}{2}$$.

• Thank you very much! You really helped me Nov 11, 2023 at 9:49
• Looks way overkill. Also how do you compute the Taylor series of $f$? Nov 11, 2023 at 19:26
• @Adayah: you only need a few terms of the Taylor series, so it is quite easy (it is similar to what Sine of the Time did). You can also compute the derivative of $f$ using usual formulas if you prefer. Nov 12, 2023 at 3:15

$$\lim_{x \to \infty} (x^3+6x^2+1)^{\frac13}-(x^2+x+1)^{\frac12}$$ Using $$(1+t)^a\sim1+at$$ as $$t\to0$$, you have: $$\sqrt[3]{x^3+6x^2+1}-\sqrt{x^2+x+1}=x\sqrt[3]{1+6/x+1/x^3}-x\sqrt{1+1/x+1/x^2}\sim \\ x(1+2/x)-x(1+1/2x)=2-1/2=3/2$$

• What exactly do you mean by $\sim$ ? Nov 11, 2023 at 19:28
• @Adayah asymptotic equivalence Nov 11, 2023 at 19:28
• If you mean $f \sim g \iff \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1$, then the argument is incorrect for multiple reasons. A typical example of how this fails: $(x+1) - x \sim x - x = 0$, so $\lim_{x \to \infty} (x+1) - x = 0$. Nov 11, 2023 at 19:37
• @Adayah (x+1)-x =1 so it doesn't make sense saying $\sim 0$ Nov 11, 2023 at 19:51
• It doesn't, right? But I only followed the same rules that you did. How do you rigorously justify this? $$x \sqrt[3]{1+6/x+1/x^3} - x \sqrt{1+1/x+1/x^2} \sim x(1+2/x) - x(1+1/2x)$$ Nov 11, 2023 at 20:32