Question Video: Finding the Limit of a Composition of Root and Rational Functions at a Point | Nagwa Question Video: Finding the Limit of a Composition of Root and Rational Functions at a Point | Nagwa

# Question Video: Finding the Limit of a Composition of Root and Rational Functions at a Point Mathematics • Second Year of Secondary School

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Find lim_(𝑥 → 11) √((𝑥² + 10𝑥 − 11)/(𝑥² + 11𝑥)).

02:38

### 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.

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