### Video Transcript

Given π of π₯ is equal to the square root of four plus π₯ squared minus two divided by eight π₯ squared, define, if possible, π evaluated at zero so that π is continuous at π₯ is equal to zero.

In this question, weβre given a function π of π₯, and we need to determine if itβs possible to make this function continuous when π₯ is equal to zero by defining a value of π evaluated at zero. To do this, letβs start by recalling how we check the continuity of a function at a point. And we recall to check a function π is continuous at π₯ is equal to π, we need to check three things are true. We need π evaluated π to be defined. This is the same as saying that π is in the domain of our function π.

Next, we need the limit as π₯ approaches π of π of π₯ to exist. And finally, we need this limit as π₯ approaches π of our function π of π₯ to be equal to π evaluated at π. In our case, we want π to be continuous when π₯ is equal to zero. So letβs set our value of π equal to zero. This then gives us the following. We can check each of these conditions for our function π of π₯ in turn. We can start by checking that π evaluated at zero is defined. However, we can already see that weβre dividing by π₯ in our function π of π₯. And we canβt divide by zero, so π evaluated at zero is not defined.

However, we are told in the question, weβre allowed to set π evaluated at zero to be any value we want. So weβll leave this for now and move on to the second condition. We want to determine if the limit as π₯ approaches zero of our function π of π₯ exists. We might be tempted to do this by looking at left and right limits. However, we can actually just evaluate the limit as π₯ approaches zero of our function π of π₯ directly. We know we canβt evaluate this limit by using direct substitution since weβre dividing by π₯. So instead, we need to notice in our numerator, we have a conjugate expression.

And we can recall if weβre asked to evaluate the limit of an algebraic expression which contains a conjugate, we can try and simplify this limit by multiplying the numerator and denominator by this conjugate. So letβs try this. Weβre going to multiply the numerator and denominator by root four plus π₯ squared plus two. We might also call this rationalizing the numerator.

If we were to multiply the two terms in the numerator, we would see that this is a factoring of a difference between squares. So we get root four plus π₯ squared all squared minus two squared which is just four plus π₯ squared minus two squared. Then, in the denominator, we just multiply the two denominators. We get eight π₯ squared multiplied by the square root of four plus π₯ squared plus two. And now, we can start simplifying. In our numerator, we have four minus two squared, which is zero. Then, we have a shared factor of π₯ squared in the numerator and denominator. So this just leaves us with the limit as π₯ approaches zero of one divided by eight multiplied by the square root of four plus π₯ squared plus two.

And now that weβve canceled the π₯ squared in our denominator, we can see this is an algebraic expression. So we come once again attempt to evaluate this by direct substitution. Substituting in π₯ is equal to zero, we get one divided by eight times root four plus zero squared plus two and four plus zero squared is equal to four, and the square root of four is two. So we get eight multiplied by two plus two, which is 32. Therefore, this limit evaluates to give us one over 32. So weβve shown the second condition is true. The limit as π₯ approaches zero of π of π₯ exists. In fact, itβs equal to one over 32.

Letβs now move on to the third condition of continuity. We need the limit as π₯ approaches zero of π of π₯ to be equal to π evaluated at zero. And weβve already seen that the left-hand side of this equation is one over 32. So for this equation to hold true, we also need π evaluated at zero to be one over 32. We can achieve this by defining π evaluated at zero to be one over 32.

So by setting this value, we allow the third condition of continuity to be true. And now, zero is in the domain of our function. So the first condition is also satisfied. Therefore, we were able to show if we set π evaluated at zero to be equal to one over 32, then the function π will be continuous at π₯ is equal to zero.