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lim[h->0][1/[h(8+h)^[1/3]]-1/[2h]]...

`lim_[h->0][1/[h(8+h)^[1/3]]-1/[2h]]`

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`lim_(h->0)[1/(h(8+h)^(1/3))-1/(2h)]`
`=lim_(h->0)[(2-(8+h)^(1/3))/(2h(8+h)^(1/3))]`
Now, when we apply `lim h->0`, it is a `0/0` from.
So, we will apply L Hopitals rule.
Differentiating both numerator and denominator,
`=lim_(h->0)[-1/3(8+h)^(-2/3)]/[2(8+h)^(1/3) + (2h(8+h)^(-2/3))/3] ` `=(-1/3(8)^(-2/3))/((2(8)^(1/3)+0)`
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