Question

Question #060c5

Answer 1

`1`

Explanation:

We can rewrite this as a fraction so that l'Hospital's rule applies:

`lim_(xrarroo)xln(1+1/x)=lim_(xrarroo)ln(1+1/x)/(1/x)`

Note that this is in the indeterminate form `0/0`, so take the derivative of the numerator and denominator.

`=lim_(xrarroo)((-1/x^2)/(1+1/x))/(-1/x^2)`

Simplifying, this becomes:

`=lim_(xrarroo)((-1/x^2)/(1+1/x))(-x^2)=lim_(xrarroo)1/(1+1/x)`

Here, we see that `1/x` will go to `0` as `x` approaches infinity, so the limit equals

`=1/(1+0)=1`

Answer 2

`lim_(x->oo)xln(1+1/x)=1`

Explanation:

Alternatively, we can use the following logarithm property:
`blog_x(a)=log_x(a^b)`

This allows us to bring the `x` as an exponent:
`lim_(x->oo)ln((1+1/x)^x)`

We can recognize the bit inside the ln function as the definition for the number `e`, so this simply evaluates to:
`ln(e)=1`