We want to evaluate the limits.
We have the limit of the square root of the quantity
9 x squared plus x minus 3 x
as x approaches infinity.
Notice for the second limit
we have x approaches negative infinity.
First notice how as x approaches to infinity
the square root of the quantity
9 x squared plus x approaches infinity.
And then, we have minus 3 x
which also approaches infinity as x approaches infinity.
So right now the limit is in the form of
infinity minus infinity, which is not equal to 0.
This is an indeterminate form.
So I'll have to perform some algebra on the function
to hopefully then determine the limit.
So what we'll do is write the expression
for the function as a fraction, with the denominator of one
and then multiply the numerator and the denominator
by the conjugate.
The conjugate is a square root of the quantity
9 x squared plus x plus 3 x.
Multiplying the conjugates.
When we multiply the two square roots,
we get 9 x squared plus x.
The sum of the next two products will be zero.
And the last product is negative
3 x times positive 3 x, which gives us
minus 9 x squared.
The denominator is the square root of the quantity
9 x squared plus x plus 3 x,
notice the numerator simplifies 9 x squared minus
9 x squared is zero.
We now have x divided by the square root
of the quantity 9 x squared plus x plus 3 x.
When we try to find the limit in this form.
We have the indeterminate form of infinity divided
by infinity.
So now we use a strategy that we used before.
We will divide every turn by the highest power of x
in the denominator, which in this case will be x.
Notice how we do have an x squared under the square root
but the square root of x squared is equal to x is actually
the absolute value of x.
But since x is approaching positive infinity
we know x is positive.
This idea will be important in just a moment.
So we're going to divide everything by x.
Simplifying, the enumerator is one
in the denominator.
When we divide the square root by x, we want to bring the x
under the square root,
and therefore we divide
by x squared under the, a square root.
So we have the square root of 9 x square divided
by x squared, plus x divided by x squared.
And then we still have
plus 3 x divided by x simplifies the 3
and now let's simplify under the square root.
We have the square root
of 9 x square divided by x squared is 9
plus x divided by x squared simplifies the one over x
and then we add plus 3.
And now we can find the limit in this form.
So the given limit is equal to the limit
of one divided by,
the sum of the square root of the quantity 9
plus one divided by x
and 3,
as x approaches to infinity.
The only thing affected by x is one divided by x
which approaches zero as x approaches to infinity
which gives us one divided
by the sum of the square root of 9 and 3.
The square root of 9 is equal to 3, giving us one divided
by the sum of 3 and 3, which is equal to one sixth.
The limit is equal to one sixth.
And now take a look at the second limit.
The only difference again is x is approaching
negative infinity.
If you focus on just the square root, we can ignore the plus
x because the 9 x squared overpowers the x as x approaches
negative infinity, as x approaches, negative infinity
for 9 x squared, we square the negative value
which becomes positive and then multiply by 9.
Which indicates as x approaches negative infinity
the square root is approaching positive infinity,
and then we have minus 3 x, 3 x approaches
negative infinity as x approaches negative infinity.
The limit is in the form
of infinity minus negative infinity
which simplifies to infinity plus infinity
and therefore the limit approaches positive infinity.
So for this limit, no extra algebra is required.
We can determine the limit approaches positive infinity
by analyzing the original function.
I hope you found this helpful.