Question

How to find `lim(ln(2x)/(2+x)) x->` infinity?

Answer 1

The limit equals `0`

Explanation:

Since we are of the form `oo/oo`, we can use L'Hospitals rule.

`L = lim_(x->oo) (2/(2x))/(1)`

`L = lim_(x->oo) 1/x`

This is now a recognizable and commonly seen limit.

`L = 0`

Hopefully this helps!

Answer 2

`lim_(xrarroo)ln(2x)/(2+x)=0`

Explanation:

Logarithmic functions grow slower than polynomial functions. Polynomial functions grow slower than exponential functions.

Since `ln(2x)` is logarithmic and `2+x` is a polynomial (it only has a degree of `1`, but it's still a polynomial), the polynomial in the denominator will grow faster than the logarithmic function in the numerator.

Thus, the denominator will outpace the numerator and as `x` goes to infinity, the limit approaches `0`.

Note what would happen if the fraction were inverted:

`lim_(xrarroo)ln(2x)/(2+x)=0`

`lim_(xrarroo)(2+x)/ln(2x)=oo`