# Video: Applying Properties of Limits

Assume that lim_(π₯ β 6) π(π₯) = 3 and lim_(π₯ β 6) π(π₯) = 8. Find lim_(π₯ β 6) β(π(π₯) β π(π₯)).

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### Video Transcript

Assume that the limit as π₯ tends to six of π of π₯ is equal to three and the limit as π₯ tends to six of π of π₯ is equal to eight. Find the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯.

We need to find the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯. We can break this limit down using the properties of limits. We have the property for the limits of roots of functions. It tells us that the limit as π₯ tends to some constant π of the πth root of some function π of π₯ is equal to the πth root of the limit as π₯ tends to π of π of π₯. The limit weβre trying to find is the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯. So we have the limit of a square root of a function. We can therefore apply our rule for limits of roots of functions. It tells us that our limit is equal to the square root of the limit as π₯ tends to six of π of π₯ minus π of π₯.

Now, we can see that we have the limit of a difference of functions since our limit is of π of π₯ minus π of π₯. We can apply the rule for the limit of differences of functions, which tells us that the limit as π₯ tends to some constant π of a difference of functions β so thatβs π of π₯ minus π of π₯ β is equal to the limit as π₯ tends to π of π of π₯ minus the limit as π₯ tends to π of π of π₯. We can apply this rule to our limit within the square root, giving us that our limit is equal to the square root of the limit as π₯ tends to six of π of π₯ minus the limit as π₯ tends to six of π of π₯.

Now, we can spot that the limits within our square root have been given to us in the question. We have that the limit as π₯ tends to six of π of π₯ is equal to three and the limit as π₯ tends to six of π of π₯ is equal to eight, giving us that our limit is equal to the square root of eight minus three. Simplifying this, we obtain our solution which is that the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯ is equal to the square root of five.