Functions, Graphs, and Limits - AP Calculus AB

Card 1 of 2277

Didn't Know

Knew It

1 of 2019 left

Question

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 2^{-} }f(x)= \lim_{x\rightarrow 2^{-} } (x^{2}- 3x+ 17) = 2^{2}- 3(2)+ 17 = 15$

$\lim_{x\rightarrow 2^{+} }f(x) = \lim_{x\rightarrow 2^{+} }(x^{2}- 4x+ 16) = 2^{2}- 4(2)+16 = 12$

$\lim_{x\rightarrow 2 }f(x)$ does not exist, because $\lim_{x\rightarrow 2^{-} }f(x) \ne \lim_{x\rightarrow 2^{+} }f(x)$ .

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 2^{-} }f(x)= \lim_{x\rightarrow 2^{-} } (x^{2}- 3x+ 17) = 2^{2}- 3(2)+ 17 = 15$

$\lim_{x\rightarrow 2^{+} }f(x) = \lim_{x\rightarrow 2^{+} }(x^{2}- 4x+ 16) = 2^{2}- 4(2)+16 = 12$

$\lim_{x\rightarrow 2 }f(x)$ does not exist, because $\lim_{x\rightarrow 2^{-} }f(x) \ne \lim_{x\rightarrow 2^{+} }f(x)$ .

← Didn't Know|Knew It →

Question

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Tap to reveal answer

Answer

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

← Didn't Know|Knew It →

Question

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Tap to reveal answer

Answer

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

← Didn't Know|Knew It →

Question

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Tap to reveal answer

Answer

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

← Didn't Know|Knew It →

Question

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Tap to reveal answer

Answer

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

← Didn't Know|Knew It →

Question

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Tap to reveal answer

Answer

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 2^{-} }f(x)= \lim_{x\rightarrow 2^{-} } (x^{2}- 3x+ 17) = 2^{2}- 3(2)+ 17 = 15$

$\lim_{x\rightarrow 2^{+} }f(x) = \lim_{x\rightarrow 2^{+} }(x^{2}- 4x+ 16) = 2^{2}- 4(2)+16 = 12$

$\lim_{x\rightarrow 2 }f(x)$ does not exist, because $\lim_{x\rightarrow 2^{-} }f(x) \ne \lim_{x\rightarrow 2^{+} }f(x)$ .

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 2^{-} }f(x)= \lim_{x\rightarrow 2^{-} } (x^{2}- 3x+ 17) = 2^{2}- 3(2)+ 17 = 15$

$\lim_{x\rightarrow 2^{+} }f(x) = \lim_{x\rightarrow 2^{+} }(x^{2}- 4x+ 16) = 2^{2}- 4(2)+16 = 12$

$\lim_{x\rightarrow 2 }f(x)$ does not exist, because $\lim_{x\rightarrow 2^{-} }f(x) \ne \lim_{x\rightarrow 2^{+} }f(x)$ .

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

← Didn't Know|Knew It →

Question

Find all vertical asymptotes and horizontal asymptotes of the function,

$f(x) = \frac{1-2x^3}{x^3-4x^2+4x}$

Tap to reveal answer

Answer

$f(x) = \frac{1-2x^3}{x^3-4x+2x}$

1) To find the horizontal asymptotes, find the limit of the function as $x->\infty$ ,

$\lim_{x->\infty}\frac{1-2x^3}{x^3-4x^2+4x}=\lim_{x->\infty}\frac{\frac{1}{x^3}-2}{1-\frac{4}{x}+\frac{4}{x^2}} = -2$

Therefore, the function has a horizontal asymptote

2) Vertical asympototes will occur at points where the function blows up, $y->\infty$ . For rational functions this behavior occurs when the denominator approaches zero.

Factor the denominator and set to zero,

$x = \left \{ 0, 2\right \}$

So the graph of has two vertical asymptotes, one at and the other at . They have been drawn into the graph of below. The blue curves represent .

← Didn't Know|Knew It →

Question

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Tap to reveal answer

Answer

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

← Didn't Know|Knew It →

Question

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Tap to reveal answer

Answer

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

← Didn't Know|Knew It →

Question

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

Tap to reveal answer

Answer

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 2^{-} }f(x)= \lim_{x\rightarrow 2^{-} } (x^{2}- 3x+ 17) = 2^{2}- 3(2)+ 17 = 15$

$\lim_{x\rightarrow 2^{+} }f(x) = \lim_{x\rightarrow 2^{+} }(x^{2}- 4x+ 16) = 2^{2}- 4(2)+16 = 12$

$\lim_{x\rightarrow 2 }f(x)$ does not exist, because $\lim_{x\rightarrow 2^{-} }f(x) \ne \lim_{x\rightarrow 2^{+} }f(x)$ .

← Didn't Know|Knew It →

Knew It

Functions, Graphs, and Limits - AP Calculus AB

Which of the following is equal to ?

Which of the following is equal to ?

Which of the following is equal to ?

Evaluate the limit:

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

Which of the following is equal to ?

Evaluate the limit:

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

Which of the following is equal to ?

Evaluate the limit:

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

Evaluate the limit:

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

Evaluate the limit:

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

Which of the following is equal to ?

Which of the following is equal to ?

Which of the following is equal to ?

Find all vertical asymptotes and horizontal asymptotes of the function,

Evaluate the limit:

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

Which of the following is equal to ?

Which of the following is equal to ?

Which of the following is equal to ?

Evaluate the limit:

To evaluate the limit, we must first determine whether the limit is right or left sided; the minus sign exponent indicates we are approaching three from the left, or using values slightly less than three. These correspond to the first part of the piecewise function, whose limit is 9 (substitution).

Which of the following is equal to ?

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Find all vertical asymptotes and horizontal asymptotes of the function,

$f(x) = \frac{1-2x^3}{x^3-4x^2+4x}$

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Evaluate the limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2, x<3\\ln(x) , x\geq 3 \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \ 17, & x=2\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?