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Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$

$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\rightarrow \infty$

So Using $\bf{A.M\geq G.M}\;,$ We get $$\frac{x+1+x+2+x+3+x+4+x+5}{5}\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$

So $$x+3\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$

So $$\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\leq 3$$

and equality hold when $x+1=x+2=x+3=x+4=x+5\;,$ Where $x\rightarrow \infty$

So $$\lim_{x\rightarrow 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3$$

Can we solve the above limit in that way, If not then how can we calculate it

and also plz explain me where i have done wrong in above method

Thanks

Martin Sleziak
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juantheron
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  • I'm not sure about the equality part. That is weird. I would suggest instead to produce a lower bound for your inequality and try to get it to reach 3. – Element118 Feb 22 '16 at 07:09
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    Let $y^5=(x+1)(x+2)(x+3)(x+4)(x+5)$. Then $y-x=\frac{y^5-x^5}{y^4+y^3x+y^2x^2+yx^3+x^4}$. The dominant term on top is $15x^4$, and on the bottom it is $5x^4$. – André Nicolas Feb 22 '16 at 07:26
  • Let $y=(x+1)(x+2)(x+3)(x+4)(x+5)$, then $y=x^5+15x^4+o(x^4)$ and $$(1+u)^{1/5}=1+u/5+o(u)$$ when $u\to0$ hence $$y^{1/5}=x(1+15/x+o(1/x))^{1/5}=x(1+3/x+o(1/x))=x+3+o(1),$$ which means exactly that $$\lim_{x\to\infty}((x+1)(x+2)(x+3)(x+4)(x+5))^{1/5}-x=3.$$ Note that the limit is actually $$(1+2+3+4+5)/5,$$ hence this systematic (and short...) method also shows that, for every $n\geqslant1$ and $(a_k)$, $$\lim_{x\to\infty}\left(\prod_{k=1}^n(x+a_k)\right)^{1/n}-x=\frac1n\sum_{k=1}^na_k.$$ – Did Feb 22 '16 at 07:42
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    You proof isn't sufficient, as your equality is one-sided only. The claim with $x+1=x+2\cdots$ lacks rigor. –  Feb 22 '16 at 07:49
  • Related (the question over there is more specific but my answer to it is more general). – Did Feb 22 '16 at 08:50

7 Answers7

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\begin{align} &\lim_{x\to\infty}\left(\left((x+1)(x+2)(x+3)(x+4)(x+5)\right)^{\frac15}-x\right)\\ &=\lim_{x\to\infty}\cfrac{\left((1+\frac1x)(1+\frac2x)(1+\frac3x)(1+\frac4x)(1+\frac5x)\right)^{\frac15}-1}{\frac1x}\\ &=\lim_{h\to0}\cfrac{\left((1+h)(1+2h)(1+3h)(1+4h)(1+5h)\right)^{\frac15}-1}{h}\\ &=f'(0)\\ f(x)&=\left((1+x)(1+2x)(1+3x)(1+4x)(1+5x)\right)^{\frac15}\\ f(x)^5&=(1+x)(1+2x)(1+3x)(1+4x)(1+5x)\\ \left(f(x)^5\right)'&=5f(x)^4f'(x)\\&=f(x)\left(\frac1{1+x}+\frac2{1+2x}+\frac3{1+3x}+\frac4{1+4x}+\frac5{1+5x}\right)\\ f'(0)&=\frac{\left.\left(f(x)^5\right)'\right|_{x=0}}{5f(0)^4}\\ &=\frac{1+2+3+4+5}{{5f(0)^3}}=\frac{1+2+3+4+5}{5}=3 \end{align}

Kay K.
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7

Let us shift the variable by $3$ and get

$$\lim_{x\to\infty}\sqrt[5]{(x-2)(x-1)x(x+1)(x+2)}-x+3=\lim_{x\to\infty}x\left(\sqrt[5]{1-\frac5{x^2}+\frac4{x^4}}-1\right)+3.$$

Then by L'Hospital,

$$\lim_{t\to0}\frac{\sqrt[5]{1-5t^2+4t^4}-1}t=\lim_{t\to0}\frac{-10t+16t^3}{5\sqrt[5]{1-5t^2+4t^4}}=0.$$

4

Hint:
Your method can be completed like this $$HM \le GM \le AM \implies$$ $$\frac5{\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}+\frac1{x+5}} \le \sqrt[5]{(x+1)(x+2)(x+3)(x+4)(x+5)} \le x+3$$ Subtracting $x$ throughout : $$\frac{15x^4+170x^3+675x^2+1096x+600}{5x^4+60x^3+255x^2+450x+274} \le \sqrt[5]{(x+1)(x+2)(x+3)(x+4)(x+5)} -x \le 3$$

Now take the limit as $x \to \infty$.

Macavity
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3

$$\lim _{t\to 0}\left(\left[\left(\frac{1}{t}+1\right)\left(\frac{1}{t}+2\right)\left(\frac{1}{t}+3\right)\left(\frac{1}{t}+4\right)\left(\frac{1}{t}+5\right)\right]^{\frac{1}{5}}-\frac{1}{t}\right) = \lim _{t\to 0}\left(\frac{\sqrt[5]{1+15t+85t^2+225t^3+274t^4+120t^5}-1}{t}\right) $$ Now we use the Taylor's development at the first order $$= \lim _{t\to 0}\left(\frac{1+3t-1+o(t)}{t}\right) = \color{red}{3}$$

Amarildo
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2

Let $t=x+3$, then $$ ((x+1)(x+2)(x+3)(x+4)(x+5))^{\frac{1}{5}}-x=(t^5-5t^3+4t)^{\frac{1}{5}}-(t-3). $$ Using $x^5-a^5=(x-a)(x^4+ax^3+a^2x^2+a^3x+a^4)$, \begin{align} &(t^5-5t^3+4t)^{\frac{1}{5}}-(t-3)\\ &=\frac{t^5-5t^3+4t-(t-3)^5}{(t^5-5t^3+4t)^{\frac{4}{5}}+(t^5-5t^3+4t)^{\frac{3}{5}}(t-3)+(t^5-5t^3+4t)^{\frac{2}{5}}(t-3)^2+(t^5-5t^3+4t)^{\frac{1}{5}}(t-3)^3+(t-3)^4}. \end{align} It looks complicated, and really it is. But we can compute limit comparing coefficients of $t^4$ with numerator and denominator. By binomial theorem, coeffecient of $t^5$ is $0$ and that of $t^4$ is $15$. If we divide numerator and denominator by $t^4$ and sends $t$ to $\infty$, numerator remains $15$. How about denominator? Denominator divided by $t^4$ is $$ (1-5t^{-2}+5t^{-4})^{\frac{4}{5}}+(1-5t^{-2}+5t^{-4})^{\frac{3}{5}}(1-3t^{-1})+(1-5t^{-2}+5t^{-4})^{\frac{2}{5}}(1-3t^{-1})^2+(1-5t^{-2}+5t^{-4})^{\frac{1}{5}}(1-3t^{-1})^3+(1-3t^{-1})^4 $$ and goes to $1+1\cdot 1 + 1\cdot 1 + 1\cdot 1 +1=5$ as $t\to \infty$. Therefore, $$ \lim_{x\to 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3. $$

choco_addicted
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2

Setting the AM $x+3=\dfrac1y$ to find

$$\lim_{y\to0}\dfrac{(1-5y^2+4y^4)^{1/5}-(1-3y)}y$$

$$=\lim_{y\to0}\dfrac{(1-5y^2+4y^4)-(1-3y)^5}y\cdot\dfrac1{\lim_{y\to0}\sum_{r=0}^4\{(1-5y^2+4y^4)^{1/5}\}^r(1-3y)^{4-r}}$$

Using Binomial expansion, this becomes

$$\lim_{y\to0}\dfrac{5\cdot3y+y^2\left(-5-\binom52\cdot3^2\right)+\cdots+(3y)^5}y\cdot\dfrac1{\lim_{y\to0}\sum_{r=0}^4\{1^{1/5}\}^r1^{4-r}}=?$$

Use $y\to0\implies y\ne0$

lab bhattacharjee
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  • The crucial step in this answer (somewhat hidden, it seems) is $(1-5y^2+4y^4)-(1-3y)^5=5\cdot3y+O(y^2)$. If using $(1+u)^a=1+au+o(u)$ is allowed, skip all the rest and use it from the start with $a=1/5$, otherwise what s the point of the algebraic manipulations in this answer? – Did Feb 22 '16 at 08:30
  • @Did, Here I've Binomial Expansion for $$(1-3y)^5$$ – lab bhattacharjee Feb 22 '16 at 08:33
2

Clearly $$(x + 1)(x + 2)(x + 3)(x + 4)(x + 5) = x^{5} + 15x^{4} + \cdots$$ and as we will show later we don't need to bother about the coefficients of $x^{3}, x^{2}, x, x^{0}$ in order to solve this problem.

Let $P(x)$ be a monic polynomial of degree $n$ and let the coefficient of $x^{n - 1}$ in $P(x)$ be $a$ so that $$P(x) = x^{n} + ax^{n - 1} + bx^{n - 2} + \cdots$$ We prove that $$\lim_{x \to \infty}\left\{\sqrt[n]{P(x)} - x\right\} = \frac{a}{n}$$ and thus for our current problem $n = 5, a = 15$ so that the desired limit is $a/n = 3$. Clearly if we set $g(x) = \sqrt[n]{P(x)}$ then we can see that $$\frac{g(x)}{x} = \sqrt[n]{\frac{P(x)}{x^{n}}} = \sqrt[n]{1 + \frac{a}{x} + \cdots } \to 1\text{ as }x \to \infty$$ and hence \begin{align} L &= \lim_{x \to \infty}\left\{\sqrt[n]{P(x)} - x\right\}\notag\\ &= \lim_{x \to \infty}\frac{P(x) - x^{n}}{g(x)^{n - 1} + xg(x)^{n - 2} + \dots + x^{n - 1}}\notag\\ &= \lim_{x \to \infty}\frac{ax^{n - 1} + bx^{n - 2} + \cdots}{g(x)^{n - 1} + xg(x)^{n - 2} + \dots + x^{n - 1}}\notag\\ &= \lim_{x \to \infty}\dfrac{a + \dfrac{b}{x} + \cdots}{\left(\dfrac{g(x)}{x}\right)^{n - 1} + \left(\dfrac{g(x)}{x}\right)^{n - 2} + \dots + \dfrac{g(x)}{x} + 1}\notag\\ &= \frac{a + 0 + \cdots}{1 + 1 + \cdots + n \text{ terms}}\notag\\ &= \frac{a}{n} \end{align}

Paramanand Singh
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