lim f(x) = ln(x) / x x->0+ To do this

lim f(x) = ln(x) / x x->0+ To do this

Lim f (x) = ln (x) / x = LIM (LN (x)) '/ X' = Lim 1 / x = infinity
x->0+ x->0+ x->0+

LIM (x tends to positive infinity) cosx / (e ^ x + e ^ - x)

X tends to positive infinity
Then cosx oscillates at [- 1,1], that is bounded
E ^ x tends to be positive infinity
If x tends to positive infinity, - x tends to negative infinity
So e ^ - x tends to zero
So the denominator tends to infinity
And molecules are bounded
So the original formula = 0

It is proved that LIM (1 / x) = 1 / x0
General results are required when x tends to x0

Let ε > 0, let | 1 / X-1 / x0 | = | x-x0 | / | x * x0|