Question #060c5

Question

4933 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-22T15:47:56

$1$ $1$

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 · 2018-01-14T16:07:13

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

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:
$ln(e)=1$