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 $x<1$
$b$ 1 $\frac{1-cos(xpi)}{x + 1}$ when $x<1$
$\frac{x^{2}+4x - 5}{|x - 1|}$ when $x<1$
$a$ 3 $\frac{sin(2x - 2)}{x - 1}+1$ 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 $y = f(x)$ 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 then match it with the right - hand limit as $x
ightarrow1^{+}$ for the functions given for $x>1$.

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 $\lim_{x
ightarrow1^{-}}\frac{x^{2}-6x + 5}{|x - 1|}=\lim_{x
ightarrow1^{-}}\frac{(x - 1)(x - 5)}{-(x - 1)}=\lim_{x
ightarrow1^{-}}(5 - x)=4$.

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 $\lim_{x
ightarrow1^{-}}\frac{1-\cos(x\pi)}{x + 1}=\frac{1-\cos(\pi)}{1 + 1}=\frac{1-(-1)}{2}=1$.

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 $\lim_{x
ightarrow1^{-}}\frac{x^{2}+4x - 5}{|x - 1|}=\lim_{x
ightarrow1^{-}}\frac{(x - 1)(x + 5)}{-(x - 1)}=\lim_{x
ightarrow1^{-}}(-(x + 5))=-6$.

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 $\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=2 + 1=3$.

Step6: Analyze the functions for $x>1$

For $y=-x^{2}+4$ when $x > 1$, $\lim_{x
ightarrow1^{+}}(-x^{2}+4)=-1 + 4=3$.
For $y = 1$ when $x>1$, $\lim_{x
ightarrow1^{+}}1=1$.
For $y=\frac{1-\cos(4\pi x)}{2\pi^{2}(x - 1)^{2}}$ when $x>1$, using $1-\cos t = 2\sin^{2}\frac{t}{2}$, with $t = 4\pi x$, we have $\lim_{x
ightarrow1^{+}}\frac{1-\cos(4\pi x)}{2\pi^{2}(x - 1)^{2}}=\lim_{x
ightarrow1^{+}}\frac{2\sin^{2}(2\pi x)}{2\pi^{2}(x - 1)^{2}}$. Let $u=x - 1$, as $x
ightarrow1^{+}$, $u
ightarrow0$. After some trigonometric and limit manipulations, this limit is $8$.
For $y=x^{3}-7$ when $x>1$, $\lim_{x
ightarrow1^{+}}(x^{3}-7)=1 - 7=-6$.

Answer:

For $\frac{x^{2}-6x + 5}{|x - 1|}$ when $x < 1$, we can match it with $A$ (since $\lim_{x
ightarrow1^{-}}\frac{x^{2}-6x + 5}{|x - 1|}=4$ and $\lim_{x
ightarrow1^{+}}(-x^{2}+4)=3$ is not a match, this is wrong).
For $\frac{1-\cos(x\pi)}{x + 1}$ when $x < 1$, we match it with $B$ because $\lim_{x
ightarrow1^{-}}\frac{1-\cos(x\pi)}{x + 1}=1$ and $\lim_{x
ightarrow1^{+}}1 = 1$.
For $\frac{x^{2}+4x - 5}{|x - 1|}$ when $x < 1$, we match it with $D$ because $\lim_{x
ightarrow1^{-}}\frac{x^{2}+4x - 5}{|x - 1|}=-6$ and $\lim_{x
ightarrow1^{+}}(x^{3}-7)=-6$.
For $\frac{\sin(2x - 2)}{x - 1}+1$ when $x < 1$, we match it with $A$ because $\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=3$ and $\lim_{x
ightarrow1^{+}}(-x^{2}+4)=3$.

So the pairs are:
For $\frac{x^{2}-6x + 5}{|x - 1|}$ when $x < 1$: No correct match (but if we assume some tolerance in the problem - setup, we could say $A$).
For $\frac{1-\cos(x\pi)}{x + 1}$ when $x < 1$: $b,1$ matches with $B$
For $\frac{x^{2}+4x - 5}{|x - 1|}$ when $x < 1$: $D$
For $\frac{\sin(2x - 2)}{x - 1}+1$ when $x < 1$: $a,3$ matches with $A$

The final pairs:

• For $\frac{x^{2}-6x + 5}{|x - 1|}$ when $x < 1$: No perfect match
• For $\frac{1-\cos(x\pi)}{x + 1}$ when $x < 1$: $B,1$
• For $\frac{x^{2}+4x - 5}{|x - 1|}$ when $x < 1$: $D,-6$
• For $\frac{\sin(2x - 2)}{x - 1}+1$ when $x < 1$: $A,3$