Evaluate the following limit: `underset( x rarr - 1/2) ( "lim")(8x^3+1)/(2x+1)`

Explanation

Given,

`(lim)_(x->−1/2)(8x^3+1)/(2x+1)`

Putting value of limit we get `0/0`

So,applying l-hospital rule,

`=(lim)_(x->−1/2)(8x^3+1)/(2x+1)`

`=(lim)_(x->−1/2)(24x^2)/(2)`

`=12x^2`

`=12(-1/2)^2`

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