( 1 + a 3 − 1 ) x 2 + ( 1 + a − 1 ) x + ( 1 + a 6 − 1 ) = 0 \left(\sqrt[3]{1+a}-1\right)x^{2}+\left(\sqrt{1+a}-1\right)x+\left(\sqrt[6]{1+a}-1\right)=0 ( 3 1 + a − 1 ) x 2 + ( 1 + a − 1 ) x + ( 6 1 + a − 1 ) = 0 α ( a ) , β ( a ) are its roots , a > − 1 \alpha(a),\ \beta(a) \text{ are its roots},\qquad a>-1 α ( a ) , β ( a ) are its roots , a > − 1 To find: lim a → 0 + α ( a ) , lim a → 0 + β ( a ) \text{To find: } \lim_{a\to0^{+}}\alpha(a),\quad \lim_{a\to0^{+}}\beta(a) To find: lim a → 0 + α ( a ) , lim a → 0 + β ( a ) ( 1 ) − 5 2 , 1 ( 2 ) − 1 2 , − 1 (1)\ -\dfrac{5}{2},\,1 \qquad (2)\ -\dfrac{1}{2},\,-1 ( 1 ) − 2 5 , 1 ( 2 ) − 2 1 , − 1 ( 3 ) − 7 2 , 2 ( 4 ) − 9 2 , 3 (3)\ -\dfrac{7}{2},\,2 \qquad (4)\ -\dfrac{9}{2},\,3 ( 3 ) − 2 7 , 2 ( 4 ) − 2 9 , 3 Plan: put t = ( 1 + a ) 1 / 6 ⇒ 1 + a = t 6 \text{Plan: put } t=(1+a)^{1/6}\ \Rightarrow\ 1+a=t^{6} Plan: put t = ( 1 + a ) 1/6 ⇒ 1 + a = t 6 all three radicals become powers of t \text{all three radicals become powers of } t all three radicals become powers of t a → 0 + ⇒ t → 1 + a\to0^{+}\ \Rightarrow\ t\to1^{+} a → 0 + ⇒ t → 1 + t = ( 1 + a ) 1 / 6 , a = t 6 − 1 t=(1+a)^{1/6},\qquad a=t^{6}-1 t = ( 1 + a ) 1/6 , a = t 6 − 1 a → 0 + ⇒ t → 1 + , t > 1 a\to0^{+}\ \Rightarrow\ t\to1^{+},\quad t>1 a → 0 + ⇒ t → 1 + , t > 1 1 + a 3 = ( 1 + a ) 1 / 3 = t 2 \sqrt[3]{1+a}=(1+a)^{1/3}=t^{2} 3 1 + a = ( 1 + a ) 1/3 = t 2 1 + a = ( 1 + a ) 1 / 2 = t 3 \sqrt{1+a}=(1+a)^{1/2}=t^{3} 1 + a = ( 1 + a ) 1/2 = t 3 1 + a 6 = ( 1 + a ) 1 / 6 = t \sqrt[6]{1+a}=(1+a)^{1/6}=t 6 1 + a = ( 1 + a ) 1/6 = t ( t 2 − 1 ) x 2 + ( t 3 − 1 ) x + ( t − 1 ) = 0 (t^{2}-1)x^{2}+(t^{3}-1)x+(t-1)=0 ( t 2 − 1 ) x 2 + ( t 3 − 1 ) x + ( t − 1 ) = 0 ( t − 1 ) ( t + 1 ) x 2 + ( t − 1 ) ( t 2 + t + 1 ) x + ( t − 1 ) = 0 (t-1)(t+1)x^{2}+(t-1)(t^{2}+t+1)x+(t-1)=0 ( t − 1 ) ( t + 1 ) x 2 + ( t − 1 ) ( t 2 + t + 1 ) x + ( t − 1 ) = 0 ( t − 1 ) [ ( t + 1 ) x 2 + ( t 2 + t + 1 ) x + 1 ] = 0 (t-1)\left[(t+1)x^{2}+(t^{2}+t+1)x+1\right]=0 ( t − 1 ) [ ( t + 1 ) x 2 + ( t 2 + t + 1 ) x + 1 ] = 0 t > 1 ⇒ t − 1 ≠ 0 ⇒ ( t + 1 ) x 2 + ( t 2 + t + 1 ) x + 1 = 0 t>1\Rightarrow t-1\neq0\ \Rightarrow\ (t+1)x^{2}+(t^{2}+t+1)x+1=0 t > 1 ⇒ t − 1 = 0 ⇒ ( t + 1 ) x 2 + ( t 2 + t + 1 ) x + 1 = 0 a → 0 + ⇔ t → 1 + : ( t + 1 ) → 2 , ( t 2 + t + 1 ) → 3 a\to0^{+}\ \Leftrightarrow\ t\to1^{+}:\quad (t+1)\to2,\quad (t^{2}+t+1)\to3 a → 0 + ⇔ t → 1 + : ( t + 1 ) → 2 , ( t 2 + t + 1 ) → 3 2 x 2 + 3 x + 1 = 0 2x^{2}+3x+1=0 2 x 2 + 3 x + 1 = 0 ( 2 x + 1 ) ( x + 1 ) = 0 ⇒ x = − 1 2 or x = − 1 (2x+1)(x+1)=0\ \Rightarrow\ x=-\dfrac{1}{2}\ \text{or}\ x=-1 ( 2 x + 1 ) ( x + 1 ) = 0 ⇒ x = − 2 1 or x = − 1 lim a → 0 + α ( a ) = − 1 2 , lim a → 0 + β ( a ) = − 1 \lim_{a\to0^{+}}\alpha(a)=-\dfrac{1}{2},\qquad \lim_{a\to0^{+}}\beta(a)=-1 lim a → 0 + α ( a ) = − 2 1 , lim a → 0 + β ( a ) = − 1 ( 2 ) − 1 2 and − 1 \boxed{(2)\ -\dfrac{1}{2}\ \text{and}\ -1} ( 2 ) − 2 1 and − 1
