Question

enter a letter and a number for each formula below so as to define a continuous function.
the letter refers to the list of equations and the number is the value of the function f at 1.
letter, number
\\(\frac{x^{2}-6x + 5}{|x - 1|}\\) when (x < 1)
\\(\frac{1-cos(xpi)}{x + 1}\\) when (x < 1)
\\(\frac{x^{2}+4x - 5}{|x - 1|}\\) when (x < 1)
\\(\frac{sin(2x - 2)}{x - 1}+1) when (x < 1)
a. (-x^{2}+4) when (x > 1)
b. (1) when (x > 1)
c. (\frac{1-cos(4pi x)}{2pi^{2}(x - 1)^{2}}) when (x > 1)
d. (x^{3}-7) when (x > 1)

Explanation:

Step1: Recall the definition of continuity

A function $f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. We need to find the left - hand limit as $x
ightarrow1^{-}$ for each of the given functions for $x < 1$ and the right - hand limit as $x
ightarrow1^{+}$ for each of the given functions for $x>1$ and match them.

Step2: Analyze the first function for $x < 1$: $\frac{x^{2}-6x + 5}{|x - 1|}$

Factor the numerator: $x^{2}-6x + 5=(x - 1)(x - 5)$. When $x<1$, $|x - 1|=-(x - 1)$. So $\lim_{x
ightarrow1^{-}}\frac{x^{2}-6x + 5}{|x - 1|}=\lim_{x
ightarrow1^{-}}\frac{(x - 1)(x - 5)}{-(x - 1)}=\lim_{x
ightarrow1^{-}}(5 - x)=4$.
For $x>1$, consider $y=-x^{2}+4$. Then $\lim_{x
ightarrow1^{+}}(-x^{2}+4)=-1 + 4=3$. This pair does not give continuity.

Step3: Analyze the second function for $x < 1$: $\frac{1-\cos(x\pi)}{x + 1}$

We know that $1-\cos t=2\sin^{2}\frac{t}{2}$. So $1-\cos(x\pi)=2\sin^{2}(\frac{x\pi}{2})$. Then $\lim_{x
ightarrow1^{-}}\frac{1-\cos(x\pi)}{x + 1}=\frac{1-\cos(\pi)}{1 + 1}=\frac{1-(-1)}{2}=1$.
For $x>1$, consider $y = 1$. Then $\lim_{x
ightarrow1^{+}}1=1$.

Step4: Analyze the third function for $x < 1$: $\frac{x^{2}+4x - 5}{|x - 1|}$

Factor the numerator: $x^{2}+4x - 5=(x - 1)(x + 5)$. When $x<1$, $|x - 1|=-(x - 1)$. So $\lim_{x
ightarrow1^{-}}\frac{x^{2}+4x - 5}{|x - 1|}=\lim_{x
ightarrow1^{-}}\frac{(x - 1)(x + 5)}{-(x - 1)}=\lim_{x
ightarrow1^{-}}(-(x + 5))=-6$.
For $x>1$, consider $y=-x^{2}+4$. Then $\lim_{x
ightarrow1^{+}}(-x^{2}+4)=3$. This pair does not give continuity.

Step5: Analyze the fourth function for $x < 1$: $\frac{\sin(2x - 2)}{x - 1}+1$

We know that $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$. Let $u = 2x-2$, as $x
ightarrow1$, $u
ightarrow0$. So $\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{\frac{2x - 2}{2}}+1=2 + 1=3$.
For $x>1$, consider $y=-x^{2}+4$. Then $\lim_{x
ightarrow1^{+}}(-x^{2}+4)=3$.

For the function $\frac{1-\cos(x\pi)}{x + 1}$ when $x < 1$ and $y = 1$ when $x>1$, the letter is B and the number is 1.
For the function $\frac{\sin(2x - 2)}{x - 1}+1$ when $x < 1$ and $y=-x^{2}+4$ when $x>1$, the letter is A and the number is 3.

Answer:

For $\frac{1-\cos(x\pi)}{x + 1}$ when $x < 1$: B, 1
For $\frac{\sin(2x - 2)}{x - 1}+1$ when $x < 1$: A, 3