1. Given:
A. [tex]$x+y=5$[/tex]
B. [tex]$x-y=-1$[/tex]
C. [tex]$x+y=3$[/tex]
2. Given that [tex]$5 \leq f(x) \leq x^2+6 x-2$[/tex]
Evaluate [tex]$\lim _{x \rightarrow 1} f(x)$[/tex].
A. 0
B. 1
C. 5
D. 7
3. Find the midpoint of the line segment P(1,2) and F
Answer (2)
The limit of f ( x ) as x approaches 1 is 5, based on the Squeeze Theorem, leading to the answer choice C. The solution for equations A and B is ( 2 , 3 ) , which does not satisfy equation C. The midpoint of the line segment from point P to point F cannot be determined due to undefined point F.
;
Solve equations A and B and find that the solution (2,3) does not satisfy equation C.
Apply the Squeeze Theorem to the inequality 5 \lim_{x \to 1} (x^2 + 6x - 2) = 5$.
Conclude that lim x → 1 f ( x ) = 5 .
Since point F is not defined, the midpoint of PF cannot be determined. The limit of f(x) as x approaches 1 is 5 .
Explanation
Problem Analysis We are given three equations: A. x + y = 5 B. x − y = − 1 C. x + y = 3 We need to determine which pair of equations has a solution. Then, we are given an inequality: 5 W e n ee d t oe v a l u a t e \lim_{x \rightarrow 1} f(x) . F ina ll y , w e n ee d t o f in d t h e mi d p o in t o f t h e l in ese g m e n t P(1,2) an d F , w h ere F$ is not defined.
Solving Equations A and B First, let's solve the system of equations A and B: A. x + y = 5 B. x − y = − 1 Adding equations A and B, we get: ( x + y ) + ( x − y ) = 5 + ( − 1 ) 2 x = 4 x = 2 Substituting x = 2 into equation A, we get: 2 + y = 5 y = 3 So the solution to equations A and B is ( x , y ) = ( 2 , 3 ) .
Checking Equation C Now, let's check if the solution ( 2 , 3 ) satisfies equation C: C. x + y = 3 Substituting x = 2 and y = 3 into equation C, we get: 2 + 3 = 5 e q 3 Since the solution ( 2 , 3 ) does not satisfy equation C, there is no common solution for equations A, B, and C.
Evaluating the Limit Next, let's evaluate the limit of f ( x ) as x approaches 1, given the inequality: 5 W ec an u se t h e Sq u eeze T h eore m t o f in d t h e l imi t o f f(x) a s x a pp ro a c h es 1. F i rs t , l e t ′ s f in d t h e l imi t o f t h e l o w er b o u n d a s x a pp ro a c h es 1 : \lim_{x \to 1} 5 = 5 N o w , l e t ′ s f in d t h e l imi t o f t h e u pp er b o u n d a s x a pp ro a c h es 1 : \lim_{x \to 1} (x^2 + 6x - 2) = (1)^2 + 6(1) - 2 = 1 + 6 - 2 = 5 S in ce b o t h t h e l o w er an d u pp er b o u n d s ha v e t h es am e l imi t a s x a pp ro a c h es 1 , b y t h e Sq u eeze T h eore m , w e ha v e : \lim_{x \to 1} f(x) = 5$
Midpoint of PF Finally, we need to find the midpoint of the line segment P ( 1 , 2 ) and F . However, point F is not defined in the problem. Therefore, we cannot determine the midpoint of PF .
Final Answer The limit of f ( x ) as x approaches 1 is 5. Therefore, the answer to the second part of the question is C. 5. Since point F is not defined, we cannot find the midpoint of PF.
Examples
The Squeeze Theorem is a powerful tool in calculus that helps us find the limit of a function when it is bounded between two other functions whose limits are known. In real life, this can be used to estimate values in situations where direct measurement is difficult. For example, imagine estimating the temperature inside a sealed container. If you know the temperature outside the container is always between two specific values, you can use the Squeeze Theorem to infer that the temperature inside the container will approach a value within that range as well.
