Evaluate `lim_(xtooo) (log_(e)x)/(x)`

1.	Evaluate `lim_(xtooo) (log_(e)x)/(x)`
Answer» We have `underset(xtooo)lim(log_(e)x)/(x)` When `xtooo, log_(e)xtooo.` So, we have indeterminate form `(oo)/(oo).` But `log_(e)x` is very sluggish function, i.e., increases very slowly compared to x. So, at ifinity though both `log_(e)x` and x approaches to infinity but the cifference between these infinity values is also infinity, i.e., x is far more infinite than `log_(e)x.` Therefore `underset(xtooo)lim(log_(e)x)/(x)=0`

Discussion

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