# Video: Finding the Unknown in a Piecewise-Defined Function That Makes It Continuous at a Point

Find the value of π which makes π continuous at π₯ = 3, given π(π₯) = (π₯β»ΒΉ β 3β»ΒΉ)/(π₯Β² β 3Β²) if π₯ β  3 and π(π₯) = π if π₯ = 3.

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### Video Transcript

Find the value of π which makes π continuous at π₯ equals three. Given πof π₯ equals π₯ to the minus one minus three to the minus one over π₯ squared minus three squared, if π₯ is not equal to three and π, if π₯ equals three.

By definition, the function π is continuous at a number π, if the limit of πof π₯ as π₯ approaches π is equal to π of π. We want to make our function π continuous at π₯ equals three. So with the function π defined as above, we need the limit of π of π₯, as π₯ approaches three, to be π of three. We can just read off the value of π of three from the definition of the function. π of π₯ is π if π₯ is three. So this π of three is π. Swapping the sides of the equation, we see that the value of π which makes π continuous at π₯ equals three, is the limit of π of π₯ as π₯ approaches three.

So our task becomes to find the value of this limit. Note that if this limit doesnβt exist for some reason, then thereβs not much we can do. Thereβs no value of π which will make π continuous. And so, really, weβve just got a hope that this limit does exist. Letβs find this limit then. The limit of the function, as π₯ approaches three, doesnβt depend on the value of the function at π₯ equals three, just the values near π₯ equals three. So we can replace π of π₯ inside this limit by the rule for π of π₯ when π₯ is not equal to three.

This isnβt quite written in the form of a rational function. Notice the negative exponent in the power of π₯ in the numerator which means that the numerator is not a polynomial. However, we could still hope that this function was continuous. And that we could evaluate the limit here, using direct substitution. Unfortunately, substituting three for π₯ gives zero over zero, the indeterminate form. And so, itβs not that simple. Weβre going to have to do some algebra.

The first thing we can do is to turn the powers with the exponents negative one into fractions. So π₯ to the negative one becomes one over π₯. And three to the negative one becomes one over three, or a third. And in the same step, we can factor the denominator which we recognize as a difference of two squares. We have fractions in our fractions. And we can get rid of them by multiplying both numerator and denominator by three π₯. The denominator is easy because itβs factored. We just stick the three π₯ in front with the numerator we have, three π₯ times one over π₯, which is three, minus three π₯ times a third, which is π₯.

Now, if we pull out a factor of negative one from the numerator, we get negative one times π₯ minus three. And writing the numerator in this way allows us to see a common factor which we cancel. Having cancelled the common factor of π₯ minus three in the numerator and denominator, weβre hopeful that directly substituting in π₯ equals three will give us something which is defined, and not the indeterminate form zero over zero.

Letβs try it. We substitute three for π₯. Doing so, we get negative one over three times three times three plus three, which is negative one over 54. If we define the function π by the rule π of π₯ equals π₯ to the negative one minus three to the negative one over π₯ squared minus three squared, then π is not continuous at π₯ equals three. This is because the function π is not defined for an input of three. Trying to evaluate π of three using the rule gives the indeterminate form zero over zero.

But weβve seen that we can define the function π, where π of π₯ is just π of π₯, if π₯ is not equal to three and π of π₯ is negative one over 54, if π₯ equals three. And this function is continuous at π₯ equals three. By defining π of π₯ to be negative one over 54 when π₯ is three, we managed to patch up the discontinuity in π. When we can do this, we say that the discontinuity is removable. And so we see that π has a removable discontinuity at π₯ equals three.