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lim(x->2)[(x+2)^[1/2]-(15x+2)^[1/5]]/[(7...

`lim_(x->2)[(x+2)^[1/2]-(15x+2)^[1/5]]/[(7x+2)^[1/4]-x]`

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`Lim_(x->2) ((x+2)^(1/2) - (15x+2)^(1/5))/((7x+2)^(1/4) - x)`
When we put `x = 2`, it becomes a `0/0` form. So, we will apply L`'`Hospital rule.
`= Lim_(x->2) ((x+2)^(1/2-1) - (15x+2)^(1/5-1)(15))/((7x+2)^(1/4-1)(7) - 1)`
`=((4)^(-1/2) - (32)^(-4/5)(15))/((16)^(-3/4)(7) -1)`
`= (1/2-15/16)/(7/8 -1)`
`=(-7/16)/(-1/8)`
`=7/2`
`:. Lim_(x->2) ((x+2)^(1/2) - (15x+2)^(1/5))/((7x+2)^(1/4) - x)= 7/2`
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