Question Video: Discussing the Addition Rule for Limits | Nagwa Question Video: Discussing the Addition Rule for Limits | Nagwa

# Question Video: Discussing the Addition Rule for Limits Mathematics • Second Year of Secondary School

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True or False: Given that lim_(π₯ β π) (π(π₯)) = πΏ, and lim_(π₯ β π) (π(π₯)) = πΎ, then lim_(π₯ β π) (π(π₯) + π(π₯)) = lim_(π₯ β π) (π(π₯)) + lim_(π₯ β π) (π(π₯)).

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### Video Transcript

True or False: Given that the limit as π₯ approaches π of π of π₯ is equal to πΏ and the limit as π₯ approaches π of π of π₯ is equal to πΎ, then the limit as π₯ approaches π of π of π₯ plus π of π₯ is equal to the limit as π₯ approaches π of π of π₯ plus the limit as π₯ approaches π of π of π₯.

In this question, we need to determine whether a statement is true or false. We need to determine if the limit of a function π of π₯ as π₯ approaches π exists and is equal to a finite value of πΏ and the limit as π₯ approaches π of a function π of π₯ exists and is equal to a finite value of πΎ, then is it true that the limit as π₯ approaches π of the sum of these functions π of π₯ plus π of π₯ is equal to the sum of their individual limits, thatβs πΏ plus πΎ? And we can answer this question directly by recalling one of the properties or algebras of limits. This is actually just the addition rule for limits. So the answer to this question is true.

However, itβs not very fulfilling to just answer a question by recalling a result. And the proof of this statement is beyond the scope of this video. But there is one interesting thing we can ask. Why do we need the condition that the limit as π₯ approaches π of π of π₯ exists and is equal to πΏ and the limit as π₯ approaches π of π of π₯ exists and is equal to πΎ? And thereβs a few different reasons we can come up with for these conditions. First, it doesnβt really make sense to say that a limit is equal to the sum of two limits if one of these limits doesnβt exist. However, this line of thinking can give us an interesting idea. If one of these two limits doesnβt exist, is that enough to prove that the limit as π₯ approaches π of π of π₯ plus π of π₯ does not exist as well?

And the answer to this question is surprisingly no. To show this, letβs consider the functions π of π₯ is one over π₯, π of π₯ is negative one over π₯, and π is zero. If π of π₯ is one over π₯ and π of π₯ is negative one over π₯ and π is zero, the left-hand side of this equation simplifies to give us the limit as π₯ approaches zero of one over π₯ minus one over π₯. And one over π₯ minus one over π₯ is equal to zero. Remember, weβre taking a limit, so weβre not worried about the case when π₯ is equal to zero. This just gives us the limit as π₯ approaches zero of zero, which is just equal to zero.

However, if we look at the individual limits of each of these functions as π₯ approaches zero, we can notice that both of these do not exist. In particular, as π₯ approaches zero, both of these limits are unbounded. And this tells us the limit of the sum of two functions is only dependent on the sum of the limit of the two functions if the limit of each function exists. But to answer this question, itβs true if the limit as π₯ approaches π of π of π₯ is equal to πΏ and the limit as π₯ approaches π of π of π₯ is equal to πΎ, then the limit as π₯ approaches π of π of π₯ plus π of π₯ is equal to the limit as π₯ approaches π of π of π₯ added to the limit as π₯ approaches π of π of π₯.

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