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Question Number 146529 by 0731619 last updated on 13/Jul/21

Answered by Olaf_Thorendsen last updated on 14/Jul/21

$$\mathrm{Let}x=5+ϵ$$$$\frac{x!!!-10}{x-5}=\frac{(5+ϵ)(2+ϵ)-10}{5+ϵ-5}$$$$=\frac{7ϵ+{ϵ}^2}{ϵ}\underset{ϵ\to 0}{\sim }7$$

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