Question

What is the limit of `sqrt(10x^2 -7x +6)/(5x + 8)` as x approaches infinity?

Answer 1

`sqrt(2/5)`

Explanation:

In the bottom fraction `5x+8` will start to look more like `5x` as `x` gets very big.

At the same time `sqrt(10x^2-7x+6)` will start to look like `sqrt(10x^2) = sqrt(10)x` as the `x^2` term dominates the others.

Thus `sqrt(10x^2-7x+6)/(5x+8)->(sqrt(10)x)/(5x)` as `x-> oo`

`(sqrt(10)x)/(5x)=sqrt(10)/5=(sqrt2sqrt(5))/5=sqrt(2/5)`

Hence our final answer: Here is a graph of the function. As you can see it asymptotically approaches `sqrt(2/5)`

We also see that as `x` goes to negative infinity the function tends towards `-sqrt(2/5)`.

Answer 2

`lim_(xrarroo)sqrt(10x^2-7x+6)/(5x+8) = sqrt10/5`

Explanation:

For all `x != 0`, we have

`sqrt(10x^2-7x+6) = sqrt(x^2)sqrt(10-7/x+6/x^2)`

Furthermore, `sqrt(x^2) = abs x`.

When looking for a limit as `x` increses without bound, we are concerned only with positive values of `x`, so

`lim_(xrarroo)sqrt(10x^2-7x+6)/(5x+8) = lim_(xrarroo)(sqrt(x^2)sqrt(10-7/x+6/x^2))/(x(5+8/x))`

`= lim_(xrarroo)(cancel(x)sqrt(10-7/x+6/x^2))/(cancel(x)(5+8/x))`

`= sqrt10/5`

Bonus

If we look for the limit as `x` decreases without bound, then we are concerned with negative values of `x`. For those values we have

For `x < 0`, `sqrt(x^2) = abs x=-x`.

Therefore,

`lim_(xrarr-oo)sqrt(10x^2-7x+6)/(5x+8) = lim_(xrarr-oo)(sqrt(x^2)sqrt(10-7/x+6/x^2))/(x(5+8/x))`

`= lim_(xrarr-oo)(-cancel(x)sqrt(10-7/x+6/x^2))/(cancel(x)(5+8/x))`

`= -sqrt10/5`