Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$ 
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
 A: 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,
$$\lim_{t\to0}\frac{\sqrt[5]{1-5t^2+4t^4}-1}t=\lim_{t\to0}\frac{-10t+16t^3}{5\sqrt[5]{1-5t^2+4t^4}}=0.$$
A: 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 $x \to \infty$.
A: $$\lim _{t\to 0}\left(\left[\left(\frac{1}{t}+1\right)\left(\frac{1}{t}+2\right)\left(\frac{1}{t}+3\right)\left(\frac{1}{t}+4\right)\left(\frac{1}{t}+5\right)\right]^{\frac{1}{5}}-\frac{1}{t}\right) = \lim _{t\to 0}\left(\frac{\sqrt[5]{1+15t+85t^2+225t^3+274t^4+120t^5}-1}{t}\right) $$
Now we use the Taylor's development at the first order
$$= \lim _{t\to 0}\left(\frac{1+3t-1+o(t)}{t}\right) = \color{red}{3}$$
A: 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,
$$
\lim_{x\to 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3.
$$
A: 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 $y\to0\implies y\ne0$
A: 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}
A: \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}
