Question Video: Finding the Limit of Functions Involving Square Roots Using Rationalisation | Nagwa Question Video: Finding the Limit of Functions Involving Square Roots Using Rationalisation | Nagwa

# Question Video: Finding the Limit of Functions Involving Square Roots Using Rationalisation Mathematics • Second Year of Secondary School

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Find lim_(π₯ β 0) (β(π₯ + 9) β β(βπ₯ + 9))/π₯.

03:01

### Video Transcript

Find the limit of square root π₯ plus nine minus square root minus π₯ plus nine over π₯ as π₯ approaches zero.

We write out the limit again. We see that as the denominator is π₯, we canβt just directly substitute. Similarly, we canβt write the limit of this quotient as the quotient of the limits, as this rule only applies when the value of the limit in the denominator is nonzero.

If this were a rational function with a polynomial in the numerator rather than something involving radicals as we have, then we would hope to find a factor of π₯ in the numerator to cancel with the π₯ in the denominator.

Unfortunately, we have radicals in the numerator. And there isnβt an obvious factor of π₯. Weβre going to have to create one. And we do this by multiplying both numerator and denominator by the conjugate of the numerator. The conjugate of π minus π is π plus π. So we multiply both numerator and denominator by square root π₯ plus nine plus square root negative π₯ plus nine.

We can then simplify the numerator either by using the difference of two squares identity or by expanding and then cancelling. And we can simplify further the square root of π₯ plus nine squared is just π₯ plus nine. And the other term, square root minus π₯ plus nine squared, is just minus π₯ plus nine.

Again, we leave the denominator as it is and simplify the numerator, this time collecting like terms. π₯ minus minus π₯ gives two π₯. And nine minus nine gives nothing. So the numerator is just two π₯. And now we notice a common factor of π₯ in the numerator and denominator, which we cancel.

And now weβre at the point where we can directly substitute. Substituting zero for π₯, we get something. We can evaluate either by hand or by a calculator. Itβs two over six, or one-third. The main trick to solving this problem was to multiply by the conjugate of the numerator in order to rationalize the numerator.

After doing some algebra, this allowed us to find a factor of π₯ in the numerator, which we could then cancel with the factor in the denominator. Of course, doing this changes the domain of the function inside the limit. The original function was undefined at π₯ equals zero. But for all other values of π₯, these two functions, the original function and the simplified function, have the same output.

And as the limit as π₯ tends to zero only depends on values of π₯ near zero and not π₯ equals zero itself, the two limits are equal. The penultimate step where we directly substituted zero for π₯ could be justified using the laws of limits.

The limit of a quotient is a quotient of the limit. The limit of a sum is the sum of the limit, and so on. But at this stage, thereβs probably no need to explicitly justify all these steps unless youβre asked to in the question. As long as you understand, you can do it in principle.

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