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Evaluate the following limit : limy→0[5y3+8y23y4-16y2] - Mathematics and Statistics

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Question

Evaluate the following limit :

`lim_(y -> 0)[(5y^3 + 8y^2)/(3y^4 - 16y^2)]`

Sum
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Solution

`lim_(y -> 0)(5y^3 + 8y^2)/(3y^4 - 16y^2)`

= `lim_(y -> 0) (y^2(5y + 8))/(y^2(3y^2 - 16))`

= `lim_(y -> 0) (5y + 8)/(3y^2 - 16)    ...[(because  y -> 0","  y ≠ 0),(therefore y^2 ≠ 0)]`

= `(lim_(y -> 0) (5y + 8))/(lim_(y -> 0) (3y^2 - 16))`

= `(5 xx 0 + 8)/(3 xx 0 - 16)`

= `-1/2`.

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Chapter 7: Limits - Exercise 7.2 [Page 141]

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