### Video Transcript

Find the limit as 𝑥 approaches 11
of the square root of 𝑥 squared plus 10𝑥 minus 11 all divided by 𝑥 squared plus
11𝑥.

We’re asked to evaluate the limit
as 𝑥 approaches 11 of a function. And the first question we should
always ask ourselves is, are we allowed to evaluate this limit by using direct
substitution? In this case, the function inside
our limit is the square root of the quotient of two polynomials. First, we recall we’re allowed to
evaluate the limit of a rational function by using direct substitution. This is because it’s the quotient
of two polynomials. Therefore, our inner rational
function is found to be evaluated by using direct substitution.

Next, we need to recall we’re
allowed to evaluate the limit of any power function by using direct
substitution. And power functions are functions
in the form 𝑥 to the power of 𝑝 for some value of 𝑝. But remember, the square root of 𝑥
can be rewritten as 𝑥 to the power of one-half. So, in fact, we can see the limit
given to us in the question is the composition of functions which can be evaluated
by using direct substitution. It’s the composition of a power
function and a rational function. Therefore, we’re allowed to attempt
to evaluate this limit by using direct substitution.

So we’ll try evaluating our limit
by direct substitution. We need to substitute 𝑥 is equal
to 11 into our function. Doing this gives us the square root
of 11 squared plus 10 times 11 minus 11 all divided by 11 squared plus 11 times
11. And we can start evaluating
this. First, in the numerator inside of
our square root, we get 220. Next, if we calculate the
denominator inside of our square root, we get 242. This gives us the square root of
220 divided by 242. However, we can simplify this even
further.

Both our numerator and our
denominator inside of this square root share a factor of 22. Canceling this shared factor gives
us the square root of 10 divided by 11. And we could leave our select
this. However, we can also rationalize
our denominator. First by using our laws of
exponents, instead of taking the square root of our entire fraction, we’ll take the
square root of the numerator and the denominator separately. This gives us the square root of 10
divided by the square root of 11.

Next, to rationalize our
denominator, we’ll multiply both our numerator and our denominator through by root
11. And by doing this in our numerator,
we get the square root of 110 and in our denominator, we get 11, giving us our final
answer of the square root of 110 divided by 11. Therefore, by showing we were
allowed to evaluate our limit by direct substitution, we were able to show the limit
as 𝑥 approaches 11 of the square root of 𝑥 squared plus 10 𝑥 minus 11 all divided
by 𝑥 squared plus 11𝑥 is equal to the square root of 110 divided by 11.