Abstract In the article, we prove that the double inequality x 2 + p 0 x + p 0 < Γ ( x + 1 ) < x 2 + 9 / 5 x + 9 / 5 $$ \frac{x^{2}+p_{0}}{x+p_{0}}< \Gamma(x+1)< \frac{x^{2}+9/5}{x+9/5} $$ holds for all x ∈ ( 0 , 1 ) $x\in(0, 1)$ , we present the best possible constants λ and μ such that λ ( x 2 + 9 / 5 ) x + 9 / 5 ≤ Γ ( x + 1 ) ≤ μ ( x 2 + p 0 ) x + p 0 $$ \frac{\lambda(x^{2}+9/5)}{x+9/5}\leq\Gamma(x+1)\leq\frac{\mu (x^{2}+p_{0})}{x+p_{0}} $$ for all x ∈ ( 0 , 1 ) $x\in(0, 1)$ , and we find the value of x ∗ $x^{\ast}$ in the interval ( 0 , 1 ) $(0, 1)$ such that Γ ( x + 1 ) > ( x 2 + 1 / γ ) / ( x + 1 / γ ) $\Gamma(x+1)>(x^{2}+1/\gamma)/(x+1/\gamma)$ for x ∈ ( 0 , x ∗ ) $x\in(0, x^{\ast})$ and Γ ( x + 1 ) < ( x 2 + 1 / γ ) / ( x + 1 / γ ) $\Gamma(x+1)
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