Suppose

Question

Suppose

\lim_{x \to a} f (x) = L

and

\lim_{x \to a} g (x) = M

. Prove that

\lim_{x \to a} (f (x) - g (x)) = L - M

.

Accepted Answer

Hi, everyone, let's take a look at this practice problem dealing with limits. This problem says let the limit as X purchase A of G of X is equal to Q and the limit as X of purchase A of H X equals R. Using the precise definition of limits, prove that the limit as X approaches A of the quantity of G of X minus H of X is equal to q minus R. So this question asks us to use the precise definition of limits, and we need to prove that the limit as X approaches A of the quantity of G of X minus H of X is equal to q minus R. So we're actually gonna start off by writing the precise definition for that limit. And so, the definition, the precise definition says that for every epsilon, greater than 0, there exists a delta greater than 0. That's that. If 0 is less than the absolute value of X minus A, which is less than delta, Then it follows That the absolute value of G of X. Minus H X Minus the quantity of minus R in quantity has to be less than epsilon. So, we want to do is actually prove that this statement is true, using the other two limits that we are given in the problem. The we start off by looking at the precise definition of the limit, as X approaches A of G of X is equal to Q. So, that definition says that for every epsilon 1 greater than 0, there exists a delta 1 greater than 0, such that, If 0 is less than the absolute value of X minus A, which is less than delta 1, Then it follows that the absolute value of G of X minus Q. Has to be less than Epsilon one. We're also going to write out the precise definition of the limit, as X purchase A of HF X is equal to R. And so, for its definition, we will have that for every epsilon 2 greater than 0, there exists a delta 2 greater than 0. That's that. If 0 is less than the absolute value of X minus A, which is less than 22, Then it follows. That the absolute value of H of X. Minus R is less than Epsilon 2. So, we're going to use these two precise definitions to prove the definition of the other limit is true. The first thing we need to do is look at Our value of delta in our original definition. So, in order for The absolute value of X minus A to be less than 1, and the absolute value of X minus A to be less than 22, we need to choose. Delta, since it also has to be greater than the absolute value of X minus A, To be the minimum of delta 1 and Delta 2. So that, if it's the minimum, then all of the inequalities will be satisfied. So, we're going to let Delta be equal to the minimum. Of Delta one. And Delta 2. That means that all of these inequalities will be satisfied. So that takes care of the delta portion of our definition of the limit. So now we need to actually look at Epsilon. And for this, we're going to Look at the absolute value of G of X minus H X minus the quantity of Q minus R in quantity. And we're actually going to rearrange those terms. So, we're going to have the absolute value of the quantity of G of X. Minus Q in quantity minus the quantity of H of X. Minus R In quantity And here we're actually going to use the triangle inequality. So we call for the triangle inequality, that is that the absolute value of A minus B has to be less than or equal to the absolute value of A plus the absolute value of B. So, applying the triangle inequality to our expression here, for the absolute value, that means that that absolute value has to be less than or equal to the absolute value of G of X. Minus Q, which is our first term inside the absolute value signs, plus the absolute value of H of X. Minus R, which is the second term inside our absolute value signs. However, if you look at our definitions for our other two limits, with the G of X and H vex, we see that the absolute value of G of X minus Q has to be less than epsilon 1, and that the absolute value of H X minus R has to be less than epsilon 2. So that means that the sum of these two inequalities has to be less than Epsilon 1 plus Epsilon 2. So we actually want to rewrite Epsilon 1 in terms and Epsilon2 in terms of Epsilon. So what we're going to do is let Epsilon 1 equal Epsilon 2, which is going to be equal to Epsilon divided by 2. That means that Epsilon 1 plus Epsilon 2 is going to be equal to. Epsilon divided by 2, plus Epsilon divided by 2, which when we add those together, is equal to Epsilon. So what we have shown here is that the absolute value of G of X. Minus HF X. Minus the quantity of Q minus R in quantity is less than epsilon. And so that is actually what we had set out to prove. And so that means that our precise definition for the limit as X approaches A of the quantity of G of X minus H of X is equal to Q minus R is true. And we had proven this true by using the definitions, the precise definitions of the other two limits in our problem, where we have set delta to be equal to the minimum of Delta 1 and Delta 2, and Epsilon 1 equal to Epsilon 2, which is equal to epsilon divided by 2 in those definitions. So I hope that this explanation has been useful, and I'll see you in the next video.

Suppose $\lim_{x \to a} f (x) = L$ and $\lim_{x \to a} g (x) = M$ . Prove that $\lim_{x \to a} (f (x) - g (x)) = L - M$ .

Key Concepts

Limit of a Function

Limit Properties

Proof Techniques in Calculus

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