QuestionFind $\lim _{x \rightarrow 1} f(x)$ for $f(x)=\frac{x^{2}-1}{\sqrt{x}-1}$ . Choices: (A) 4, (B) 2, (C) 0, (D) nonexistent.

Studdy Solution

STEP 1

Assumptions1. The function is defined as $f(x)=\frac{x^{}-1}{\sqrt{x}-1}$ . We need to find the limit as $x$ approaches1, i.e., $\lim{x \rightarrow1} f(x)$

STEP 2

First, we will try to substitute $x=1$ into the function and see if it's defined.
$f(1)=\frac{1^{2}-1}{\sqrt{1}-1}$

STEP 3

Calculate the value of $f(1)$ .
$f(1)=\frac{1-1}{1-1}$

STEP 4

We see that the denominator becomes zero, which means that the function is undefined at $x=1$ . So, we cannot directly substitute $x=1$ into the function to find the limit. We need to simplify the function first.

STEP 5

We can simplify the function by factoring the numerator.
$f(x)=\frac{(x-1)(x+1)}{\sqrt{x}-1}$

STEP 6

We can multiply the denominator and the numerator by the conjugate of the denominator to eliminate the square root.
$f(x)=\frac{(x-1)(x+1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}$

STEP 7

implify the denominator using the difference of squares formula.
$f(x)=\frac{(x-1)(x+1)(\sqrt{x}+1)}{x-1}$

STEP 8

Cancel the common factor $(x-1)$ from the numerator and the denominator.
$f(x)=(x+1)(\sqrt{x}+1)$

STEP 9

Now, we can substitute $x=$ into the simplified function to find the limit.
$\lim{x \rightarrow} f(x) = (+)(\sqrt{}+)$

STEP 10

Calculate the limit.
$\lim{x \rightarrow} f(x) = (2)(2) =4$ So, the limit of the function as $x$ approaches is4.

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QuestionFind $\lim _{x \rightarrow 1} f(x)$ for $f(x)=\frac{x^{2}-1}{\sqrt{x}-1}$ . Choices: (A) 4, (B) 2, (C) 0, (D) nonexistent.

Studdy Solution

STEP 1

Assumptions1. The function is defined as $f(x)=\frac{x^{}-1}{\sqrt{x}-1}$ . We need to find the limit as $x$ approaches1, i.e., $\lim{x \rightarrow1} f(x)$

STEP 2

First, we will try to substitute $x=1$ into the function and see if it's defined.
$f(1)=\frac{1^{2}-1}{\sqrt{1}-1}$

STEP 3

Calculate the value of $f(1)$ .
$f(1)=\frac{1-1}{1-1}$

STEP 4

We see that the denominator becomes zero, which means that the function is undefined at $x=1$ . So, we cannot directly substitute $x=1$ into the function to find the limit. We need to simplify the function first.

STEP 5

We can simplify the function by factoring the numerator.
$f(x)=\frac{(x-1)(x+1)}{\sqrt{x}-1}$

STEP 6

We can multiply the denominator and the numerator by the conjugate of the denominator to eliminate the square root.
$f(x)=\frac{(x-1)(x+1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}$

STEP 7

implify the denominator using the difference of squares formula.
$f(x)=\frac{(x-1)(x+1)(\sqrt{x}+1)}{x-1}$

STEP 8

Cancel the common factor $(x-1)$ from the numerator and the denominator.
$f(x)=(x+1)(\sqrt{x}+1)$

STEP 9

Now, we can substitute $x=$ into the simplified function to find the limit.
$\lim{x \rightarrow} f(x) = (+)(\sqrt{}+)$

STEP 10

Calculate the limit.
$\lim{x \rightarrow} f(x) = (2)(2) =4$ So, the limit of the function as $x$ approaches is4.

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QuestionFind lim⁡x→1f(x)\lim _{x \rightarrow 1} f(x)limx→1​f(x) for f(x)=x2−1x−1f(x)=\frac{x^{2}-1}{\sqrt{x}-1}f(x)=x​−1x2−1​. Choices: (A) 4, (B) 2, (C) 0, (D) nonexistent.

Studdy Solution

STEP 1

Assumptions1. The function is defined as f(x)=x−1x−1f(x)=\frac{x^{}-1}{\sqrt{x}-1}f(x)=x​−1x−1​ . We need to find the limit as xxx approaches1, i.e., lim⁡x→1f(x)\lim{x \rightarrow1} f(x)limx→1f(x)

STEP 2

First, we will try to substitute x=1x=1x=1 into the function and see if it's defined.f(1)=12−11−1f(1)=\frac{1^{2}-1}{\sqrt{1}-1}f(1)=1​−112−1​

STEP 3

Calculate the value of f(1)f(1)f(1).f(1)=1−11−1f(1)=\frac{1-1}{1-1}f(1)=1−11−1​

STEP 4

We see that the denominator becomes zero, which means that the function is undefined at x=1x=1x=1. So, we cannot directly substitute x=1x=1x=1 into the function to find the limit. We need to simplify the function first.

STEP 5

We can simplify the function by factoring the numerator.f(x)=(x−1)(x+1)x−1f(x)=\frac{(x-1)(x+1)}{\sqrt{x}-1}f(x)=x​−1(x−1)(x+1)​

STEP 6

We can multiply the denominator and the numerator by the conjugate of the denominator to eliminate the square root.f(x)=(x−1)(x+1)(x+1)(x−1)(x+1)f(x)=\frac{(x-1)(x+1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}f(x)=(x​−1)(x​+1)(x−1)(x+1)(x​+1)​

STEP 7

implify the denominator using the difference of squares formula.f(x)=(x−1)(x+1)(x+1)x−1f(x)=\frac{(x-1)(x+1)(\sqrt{x}+1)}{x-1}f(x)=x−1(x−1)(x+1)(x​+1)​

STEP 8

Cancel the common factor (x−1)(x-1)(x−1) from the numerator and the denominator.f(x)=(x+1)(x+1)f(x)=(x+1)(\sqrt{x}+1)f(x)=(x+1)(x​+1)

STEP 9

Now, we can substitute x=x=x= into the simplified function to find the limit.lim⁡x→f(x)=(+)(+)\lim{x \rightarrow} f(x) = (+)(\sqrt{}+)limx→f(x)=(+)(​+)

STEP 10

Calculate the limit.lim⁡x→f(x)=(2)(2)=4\lim{x \rightarrow} f(x) = (2)(2) =4limx→f(x)=(2)(2)=4So, the limit of the function as xxx approaches is4.

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QuestionFind $\lim _{x \rightarrow 1} f(x)$ for $f(x)=\frac{x^{2}-1}{\sqrt{x}-1}$ . Choices: (A) 4, (B) 2, (C) 0, (D) nonexistent.

Assumptions1. The function is defined as $f(x)=\frac{x^{}-1}{\sqrt{x}-1}$ . We need to find the limit as $x$ approaches1, i.e., $\lim{x \rightarrow1} f(x)$

First, we will try to substitute $x=1$ into the function and see if it's defined.
$f(1)=\frac{1^{2}-1}{\sqrt{1}-1}$

Calculate the value of $f(1)$ .
$f(1)=\frac{1-1}{1-1}$

We see that the denominator becomes zero, which means that the function is undefined at $x=1$ . So, we cannot directly substitute $x=1$ into the function to find the limit. We need to simplify the function first.

We can simplify the function by factoring the numerator.
$f(x)=\frac{(x-1)(x+1)}{\sqrt{x}-1}$

We can multiply the denominator and the numerator by the conjugate of the denominator to eliminate the square root.
$f(x)=\frac{(x-1)(x+1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}$

implify the denominator using the difference of squares formula.
$f(x)=\frac{(x-1)(x+1)(\sqrt{x}+1)}{x-1}$

Cancel the common factor $(x-1)$ from the numerator and the denominator.
$f(x)=(x+1)(\sqrt{x}+1)$

Now, we can substitute $x=$ into the simplified function to find the limit.
$\lim{x \rightarrow} f(x) = (+)(\sqrt{}+)$

Calculate the limit.
$\lim{x \rightarrow} f(x) = (2)(2) =4$ So, the limit of the function as $x$ approaches is4.