Video: Applying Properties of Limits

Assume that lim_(π‘₯ β†’ 6) 𝑓(π‘₯) = 3 and lim_(π‘₯ β†’ 6) 𝑔(π‘₯) = 8. Find lim_(π‘₯ β†’ 6) √(𝑔(π‘₯) βˆ’ 𝑓(π‘₯)).

01:53

Video Transcript

Assume that the limit as π‘₯ tends to six of 𝑓 of π‘₯ is equal to three and the limit as π‘₯ tends to six of 𝑔 of π‘₯ is equal to eight. Find the limit as π‘₯ tends to six of the square root of 𝑔 of π‘₯ minus 𝑓 of π‘₯.

We need to find the limit as π‘₯ tends to six of the square root of 𝑔 of π‘₯ minus 𝑓 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‘Ž of the 𝑛th root of some function 𝑓 of π‘₯ is equal to the 𝑛th root of the limit as π‘₯ tends to π‘Ž of 𝑓 of π‘₯. The limit we’re trying to find is the limit as π‘₯ tends to six of the square root of 𝑔 of π‘₯ minus 𝑓 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 𝑔 of π‘₯ minus 𝑓 of π‘₯.

Now, we can see that we have the limit of a difference of functions since our limit is of 𝑔 of π‘₯ minus 𝑓 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‘Ž of a difference of functions β€” so that’s 𝑓 of π‘₯ minus 𝑔 of π‘₯ β€” is equal to the limit as π‘₯ tends to π‘Ž of 𝑓 of π‘₯ minus the limit as π‘₯ tends to π‘Ž of 𝑔 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 𝑔 of π‘₯ minus the limit as π‘₯ tends to six of 𝑓 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 𝑓 of π‘₯ is equal to three and the limit as π‘₯ tends to six of 𝑔 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 𝑔 of π‘₯ minus 𝑓 of π‘₯ is equal to the square root of five.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.