Limits, Limits, Limits, Limits

Algebra Level 3

( 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+(\sqrt [ 6 ]{ 1+a } -1)=0

Let α \alpha and β \beta be the roots of the equation (of x x ) above for a > − 1 a> - 1 .

Find the values of lim ⁡ a → 0 + α \displaystyle \lim_{a \to 0^+} \alpha and lim ⁡ a → 0 + β \displaystyle \lim_{a \to 0^+} \beta respectively.

− 9 2 , − 1 -\frac { 9 }{ 2 } , -1 − 1 2 , − 1 -\frac {1 }{ 2 } , -1 − 5 2 , − 1 -\frac { 5 }{ 2 } ,-1 − 7 2 , − 1 -\frac { 7 }{ 2 } , -1

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Ram Gautam by Ram Gautam , 22, India

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1 solution

Ram Gautam
Feb 1, 2015

G i v e n e q u a t i o n : − ( 1 + a 3 − 1 ) x 2 + ( 1 + a − 1 ) x + ( 1 + a 6 − 1 ) = 0 H e n c e , α + β = l i m a → 0 + { ( 1 + a − 1 ) ( 1 + a 3 − 1 ) } = − 3 2 α ⋅ β = l i m a → 0 + { ( 1 + a 6 − 1 ) ( 1 + a 3 − 1 ) } = 1 2 N o t e : − Y o u C a n S o l v e T h e L i m i t s B y U s i n g L − H o s p t i a l R u l e . T h e r e f o r e , B y U s i n g t h e i d e n t i t y : − ( α − β ) 2 = ( α + β ) 2 − 4 α β . W e g e t t h a t , α = − 1 2 & β = − 1 / \\ Given\quad equation:-\qquad \\ \\ \left( \sqrt [ 3 ]{ 1+a } -1 \right) { x }^{ 2 }+\left( \sqrt { 1+a } -1 \right) x+(\sqrt [ 6 ]{ 1+a } -1)=0\\ \\ Hence,\\ \alpha +\beta =\quad \underset { a\xrightarrow { } { 0 }^{ + } }{ lim } \left\{ \frac { \left( \sqrt { 1+a } -1 \right) }{ \left( \sqrt [ 3 ]{ 1+a } -1 \right) } \right\} =\frac { -3 }{ 2 } \\ \alpha \cdot \beta =\quad \underset { a\xrightarrow { } { 0 }^{ + } }{ lim } \left\{ \frac { \left( \sqrt [ 6 ]{ 1+a } -1 \right) }{ \left( \sqrt [ 3 ]{ 1+a } -1 \right) } \right\} =\frac { 1 }{ 2 } \\ Note:- \quad You\quad Can\quad Solve\quad The\quad Limits\quad By\quad Using\quad L-Hosptial\quad Rule.\\ Therefore,\\ By\quad Using\quad the\quad identity\quad :-\\ { \left( \alpha -\beta \right) }^{ 2 }={ \left( \alpha +\beta \right) }^{ 2 }-4\alpha \beta .\quad \\ We\quad get\quad that,\\ \alpha =\frac { -1 }{ 2 } \quad \& \quad \beta =-1/

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Moderator note:

Can you solve the limits without applying L'hopital rule?

You can also solve for the limits by using binominal expansion to the first order.

Prakhar Gupta - 6 years, 1 month ago

The objective approach : as L i m a → 0 + α + β = − 3 2 Lim_{a \rightarrow 0^{+} }\alpha +\beta= \frac{-3}{2} which is satisfied by the first option only. ;)

Keshav Tiwari - 6 years, 1 month ago

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Gotcha! :D

Ram Gautam - 6 years, 1 month ago

I solved this limit by using the formula for roots of quadratic equation and the substituting (1+a ) = t^6 and then factorizing the term in the square root .

Anurag Pandey - 6 years, 1 month ago

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