Question

pls help this is pretty urgent

Answer 1

Answer:

See explanation

Step-by-step explanation:

7. a. When x approaches -1, the denominator will alway be positive and close to 0. The numerator is also very close to 0. When you divide the two numbers, the answer would be 0 since the quotient is too close to 0.

b. g is f simplified.

(x^2+2x+1)/(x^2-1)=

(x+1)^2/(x+1)(x-1)=(x+1)/(x-1)

c. (1^2+2(1)+1)/(1^2-1)=

4/0.0...01 ==> ∞

When you divide a number like 4 by a very small number, the quotient gets bigger, approaching infinity.

Answer 2

Answer:

(a)  0

(b)  f(x) = g(x)

(c)  See below.

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{x^2+2x+1}{x^2-1}$

Part (a)

Factor the numerator and denominator of the given rational function:

$\begin{aligned} \implies f(x) & = \dfrac{x^2+2x+1}{x^2-1} \\\\& = \dfrac{(x+1)^2}{(x+1)(x-1)}\\\\& = \dfrac{x+1}{x-1}\end{aligned}$

Substitute x = -1 to find the limit:

$\displaystyle \lim_{x \to -1}f(x)=\dfrac{-1+1}{-1-1}=\dfrac{0}{-2}=0$

Therefore:

$\displaystyle \lim_{x \to -1}f(x)=0$

Part (b)

From part (a), we can see that the simplified function f(x) is the same as the given function g(x).  Therefore, f(x) = g(x).

Part (c)

As x = 1 is approached from the right side of 1, the numerator of the function is positive and approaches 2 whilst the denominator of the function is positive and gets smaller and smaller (approaching zero).  Therefore, the quotient approaches infinity.

$\displaystyle \lim_{x \to 1^+} f(x)=\dfrac{\to 2^+}{\to 0^+}=\infty$