Question

How do you find `\lim _ { x \rightarrow \infty } \frac { \sqrt { x ^ { 2} + x } } { 2x }`?

Answer 1

`1/2`

Explanation:

Factoring the greatest degree term from the numerator:

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

When `x>0`, as is the case as `xrarroo`, `absx=x`, so the limit becomes:

`=lim_(xrarroo)sqrt(1+1/x)/2`

As `xrarroo`, the denominator of `1/x` grows rapidly and `1/xrarr0`:

`=sqrt(1+0)/2=1/2`