## 2 Main Results  
Firstly, for simplicity's sake, we assume \(\delta_{0}(l):=\int_{l}^{\infty}a^{-1/\alpha}(\varrho)\mathrm{d}\varrho\) and \(\delta_{m}(l):=\int_{l}^{\infty}\delta_{m-1}(\varrho)\mathrm{d}\varrho\), for \(m=1,2\). Moreover, we let  
\[\mathrm{(H)}\text{ there is a constant }h>1\text{ such that }\frac{\delta_{2}(\tau(l))}{\delta_{2}(l)}\geq h \text{ for }l\geq l_{0}.\]  
When checking the behavior of positive solutions of DDE (1), we have--by using (Lemma 2.2.1, [27])--three cases:  
\[ \begin{array}{rl}\text{Case }(1):&x^{\prime}(l)>0\text{, }x^{\prime\prime \prime}(l)>0\text{ and }x^{(4)}(l)<0;\\ \text{Case }(2):&x^{\prime}(l)>0\text{, }x^{\prime\prime}(l)>0\text{ and }x^{\prime\prime \prime}(l)<0;\\ \text{Case }(3):&x^{\prime}(l)<0\text{, }x^{\prime\prime}(l)>0\text{ and }x^{\prime\prime \prime}(l)<0.\end{array} \]  
Moreover, from (1), we have that \(\big{(}a(l)\big{(}x^{\prime\prime\prime}(l)\big{)}^{\alpha}\big{)}^{\prime}\leq 0\), for \(l\in I_{1}\). We note that if \(x\in\Re\), then \(x\) satisfies Case (3). Lemma 1: _Assume that \(x\in\Re\). If_  
\[\int_{l_{0}}^{\infty}\bigg{(}\frac{1}{a(v)}\int_{l_{0}}^{v}q(\varrho)\mathrm{d }\varrho\bigg{)}^{1/\alpha}\mathrm{d}v=\infty, \tag{2}\]  
_then_  
\[\lim_{l\to\infty}x(l)=0. \tag{3}\]  
Lemma 2: _Assume that \(x\in\Re\) and (2) hold. Then_  
\[\eta:=\limsup_{l\to\infty}\delta_{2}\bigg{(}\int_{l_{0}}^{l}q(\varrho)\mathrm{ d}\varrho\bigg{)}^{1/\alpha}\leq 1. \tag{4}\]  
**Proof. ** Suppose \(x\in\Re\). Integrating (1) from \(l_{1}\) to \(l\) and using that fact that \(x^{\prime}(l)<0\), we get  
\[-a(l)\big{(}x^{\prime\prime\prime}(l)\big{)}^{\alpha} \geq -a(l_{1})\big{(}x^{\prime\prime\prime}(l_{1})\big{)}^{\alpha}+\int _{l_{1}}^{l}q(\varrho)x^{\alpha}(\tau(\varrho))\mathrm{d}\varrho \tag{5}\] \[\geq -a(l_{1})\big{(}x^{\prime\prime\prime}(l_{1})\big{)}^{\alpha}+x^{ \alpha}(\tau(l))\int_{l_{0}}^{l}q(\varrho)\mathrm{d}\varrho-x^{\alpha}(\tau(l ))\int_{l_{0}}^{l_{1}}q(\varrho)\mathrm{d}\varrho,\]  
for all \(l\in I_{1}\). In view of (3), there is a \(l_{2}\in I_{1}\) such that  
\[a(l_{1})\big{(}x^{\prime\prime\prime}(l_{1})\big{)}^{\alpha}+x^{\alpha}(\tau( l))\int_{l_{0}}^{l_{1}}q(\varrho)\mathrm{d}\varrho<0,\]  
for \(l\in I_{2}\). Thus, (5) becomes  
\[-a(l)\big{(}x^{\prime\prime\prime}(l)\big{)}^{\alpha}\geq x^{\alpha}(\tau(l)) \int_{l_{0}}^{l}q(\varrho)\mathrm{d}\varrho\geq x^{\alpha}(l)\int_{l_{0}}^{l} q(\varrho)\mathrm{d}\varrho. \tag{6}\]  
Now, by using the monotonicity of \(a^{1/\alpha}(l)x^{\prime\prime\prime}(l)\), we have  
\[x^{\prime\prime}(\tau(l))\geq x^{\prime\prime}(l)\geq\int_{l}^{\infty}\frac{1} {a^{1/\alpha}(\varrho)}\Big{(}-a^{1/\alpha}(\varrho)x^{\prime\prime\prime}( \varrho)\Big{)}\mathrm{d}\varrho\geq-a^{1/\alpha}(l)x^{\prime\prime\prime}(l) \delta_{0}(l). \tag{7}\]  
Integrating (7) twice from \(l\) to \(\infty\) and using \(\Big{(}a^{1/\alpha}(l)x^{\prime\prime\prime}(l)\Big{)}^{\prime}\leq 0\), we get  
\[-x^{\prime}(l)\geq-a^{1/\alpha}(l)x^{\prime\prime\prime}(l)\delta_{1}(l) \tag{8}\]  
and  
\[x(l)\geq-a^{1/\alpha}(l)x^{\prime\prime\prime}(l)\delta_{2}(l). \tag{9}\]  
From (9) and (6), we see that  
\[\lim_{l\to\infty}x(l)=0. \tag{3}\]0  
and so  
\[\lim_{l\to\infty}x(l)=0. \tag{3}\]1  
Taking the limsup on both sides of the inequality, we arrive at (4). The proof is complete. **Lemma 3**. : _Assume that \(x\in\Re\) and (2) hold. Then there exists a \(l_{\varepsilon}\geq l_{1}\) such that_  
\[\lim_{l\to\infty}x(l)=0. \tag{3}\]2  
_for any \(\varepsilon>0\) and \(l\in I_{\varepsilon}\). Moreover, if \((H)\) holds, then_  
\[\lim_{l\to\infty}x(l)=0. \tag{3}\]3  
**Proof. ** Suppose \(x\in\Re\). Then, there is a \(l_{1}\in I_{0}\) such that \(x(\tau(l))>0\). Proceeding as in the proof of Lemma 2, we arrive at (6) and (8). Thus, for \(l\geq l_{2}\), where \(l_{2}\in I_{1}\) is large enough, we have  
\[\lim_{l\to\infty}x(l)=0. \tag{3}\]4From the definition of \(\eta\), for every \(\varepsilon>0\), there exists a \(l_{3}\geq l_{2}\) such that  
\[\lim_{l\to\infty}x(l)=0. \tag{3}\]5  
for \(l\in I_{3}\).