Question Video: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios at a Point | Nagwa Question Video: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios at a Point | Nagwa

# Question Video: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios at a Point Mathematics • Second Year of Secondary School

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Discuss the continuity of the function π at π₯ = 0, given that π(π₯) = (6π₯Β² β sinΒ² π₯)/π₯ tan 4π₯, when π₯ < 0 and π(π₯) = (π₯ + 5)/(π₯ + 4), when π₯ β₯ 0.

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### Video Transcript

Discuss the continuity of the function π at π₯ is equal to zero, given that π of π₯ is equal to six π₯ squared minus the sin squared of π₯ divided by π₯ times the tan of four π₯ when π₯ is less than zero and π of π₯ is equal to π₯ plus five divided by π₯ plus four when π₯ is greater than or equal to zero.

In this question, weβre given a piecewise-defined function π of π₯ and we want to determine the continuity of the function π when π₯ is equal to zero. To answer this question, letβs start by recalling the three conditions which need to hold for a function to be continuous at a point. For a function π to be continuous at a value of π₯ is equal to π, we need π evaluated at π to be defined. This means π needs to be in the domain of our function π. We need the limit as π₯ approaches π of π of π₯ to exist. And we also need this limit as π₯ approaches π of π of π₯ to be equal to π evaluated at π. If all three conditions hold, then we say π is continuous at π₯ is equal to π.

In this question, weβre using the piecewise-defined function π. And we want to determine its continuity when π₯ is equal to zero. So we can substitute π is equal to zero into our definition. We now need to check all three of these conditions. Letβs start with the first condition. We want to check that zero is in the domain of our function π. And since this is a piecewise-defined function, we can see this directly from its definition. Zero is included in the second subdomain. And although we could conclude zero is in the domain of our function, itβs worth noting we need the value of π evaluated at zero for the third condition. So weβll evaluate π at zero now. We substitute π₯ is equal to zero into our second subfunction. π evaluated at zero is zero plus five divided by zero plus four, which evaluates to give us five over four. Therefore, the first condition of continuity for π at zero is true.

Letβs now move on to the second condition. We want to check that the limit as π₯ approaches zero of π of π₯ exists. And the exact method we should use to check that this function exists will depend entirely on the function weβre given and the point. In this case, we can note weβre given a piecewise-defined function and our value of π₯ is equal to zero is one of the endpoints of the subdomain. This means our function changes definition to the left of π₯ is equal to zero and to the right of π₯ is equal to zero. This should hint to us that we can check the limit as π₯ approaches zero of π of π₯ by evaluating its left and right limits.

So letβs start by evaluating the limit as π₯ approaches zero from the left of π of π₯. Since weβre taking the limit as π₯ approaches zero from the left, our values of π₯ are all going to be less than zero. And we can see when π₯ is less than zero, our function π of π₯ is exactly equal to its first subfunction. So their limits will be equal. We just need to evaluate the limit as π₯ approaches zero from the left of six π₯ squared minus the sin squared of π₯ divided by π₯ times the tan of four π₯. And we might be tempted to do this by direct substitution. However, this wonβt work. In our numerator, weβll get zero minus zero, which is equal to zero. And our denominator also has a factor of zero. So instead weβre going to need to simplify this limit and then apply our trigonometric limit results.

Letβs start by dividing both terms in the numerator separately by the denominator and splitting our limit over the difference. This gives us the limit as π₯ approaches zero from the left of six π₯ squared divided by π₯ times the tan of four π₯ minus the limit as π₯ approaches zero from the left of the sin squared of π₯ divided by π₯ multiplied by the tan of four π₯. And we can simplify this slightly. In our first limit, we can cancel the shared factor of π₯ in the numerator and denominator.

Weβre now almost ready to evaluate these limits. Weβll start by recalling that the limit as π₯ approaches zero of π₯ divided by the tan of ππ₯ is equal to one divided by π provided our value of π is not equal to zero. This will allow us to evaluate the first limit. We just need to take out the constant factor of six. We can then see that we have π₯ divided by the tan of four π₯ inside of our limit. So our value of π is equal to four. And of course, we need to multiply this by six. So we get six multiply by one-quarter. We now want to use this result along with the fact that the limit as π₯ approaches zero of the sin of ππ₯ divided by π₯ is equal to π for any real value of π to evaluate our second limit.

However, we canβt do this directly because our limit is not quite in this form. Instead, weβre going to need to multiply both the numerator and denominator of this limit by π₯. We can now see we have sin squared of π₯ in our numerator and π₯ squared in our denominator. And we also have π₯ in our numerator and tan of four π₯ in the denominator. This means we can use the productβs rule for limits to rewrite this limit as the limit as π₯ approaches zero from the left of π₯ divided by the tan of four π₯ multiplied by the limit as π₯ approaches zero from the left of sin squared of π₯ divided by π₯ squared provided both of these limits exist. And weβve already seen how to evaluate this first limit by using our trigonometric limit result. Our value of π is four, so itβs equal to one-quarter.

To evaluate our second limit however, we need to notice we have a square in our numerator and a square in our denominator. We can take the square outside of our limit by using the power rule for limits. This then gives us the limit as π₯ approaches zero from the left of the sin of π₯ divided by π₯ all squared. And now, this is our trigonometric limit result with the value of π equal to one. So the limit evaluates to give us one, and one squared is equal to one. Therefore, weβve shown this is equal to six times one-quarter minus one-quarter multiplied by one. And we can then just evaluate this. Six-quarters minus one-quarter is five-quarters. Therefore, the limit as π₯ approaches zero from the left of π of π₯ exists and itβs equal to five over four.

Letβs now clear some space and evaluate the limit as π₯ approaches zero from the right of π of π₯. This time, since weβre evaluating the limit as π₯ approaches zero from the right of π of π₯, our values of π₯ are going to be greater than zero. And we can see when π₯ is greater than zero, our function π of π₯ is exactly equal to the rational function π₯ plus five divided by π₯ plus four. Therefore, their limits as π₯ approaches zero from the right must be equal. So we just need to evaluate the limit as π₯ approaches zero from the right of π₯ plus five divided by π₯ plus four. And this is a rational function. So we can attempt to do this by using direct substitution. And we already found this when we evaluated π at zero. Itβs equal to zero plus five divided by zero plus four, which is equal to five-quarters.

Therefore, the limit as π₯ approaches zero from the right of π of π₯ exists, and itβs equal to five over four. And the limit as π₯ approaches zero from the left of π of π₯ exists, and itβs also equal to five over four. Since the left and right limit both exist and are equal, we can conclude the limit as π₯ approaches zero of π of π₯ exists and itβs equal to five over four.

And for our third and final condition, we just need to check that the limit as π₯ approaches zero of π of π₯ is equal to π evaluated at zero. We just showed the left and right limit of π of π₯ both exist and are both equal to five-quarters. So the limit as π₯ approaches zero of π of π₯ is equal to five-quarters. And weβve also already calculated π evaluated at zero; itβs also equal to five-quarters. Therefore, the third continuity condition holds, and we can conclude the function π given to us in the question is continuous at π₯ is equal to zero.

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