Question

How do you find the limit of `(3x +9) /sqrt (2x^2 +1)` as x approaches infinity?

Answer 1

`3/sqrt2`

Explanation:

Factor the largest-degreed term from the numerator and denominator.

`lim_(xrarroo)(3x+9)/sqrt(2x^2+1)=lim_(xrarroo)(x(3+9/x))/sqrt(x^2(2+1/x^2))`

Pulling the `x^2` out of the square root as just `x` (because we are only concerned with positive value of `x`), this becomes

`lim_(xrarroo)(x(3+9/x))/(xsqrt(2+1/x^2))=lim_(xrarroo)(3+9/x)/sqrt(2+1/x^2)`

Now, when we evaluate the limit as `x` approaches infinity, we see that `9/x` and `1/x^2` both go to `0`.

`=(3+0)/sqrt(2+0)=3/sqrt2`