9000062404
Level:
A
Evaluate the following limit.
\[
\lim _{x\to +\infty } \frac{x^{3} - x + 1}
{1 - x^{2} - x^{3}}
\]
\(- 1\)
\(0.5\)
\(- 0.5\)
\(1\)
Limits and continuity
9000062405
Level:
A
Evaluate the following one-sided limit.
\[
\lim _{x\to 6^{-}}\frac{3x + 2}
{x - 6}
\]
\(-\infty \)
\(1\)
\(+\infty \)
\(0\)
9000141901
Level:
A
Given the function \(f\),
find \(\lim _{x\to 1}f(x)\).
\[
f(x)=\begin{cases}
x^3+1 & \text{if } x\neq 1,\\
3 & \text{if } x = 1
\end{cases}
\]
\(2\)
\(3\)
\(1\)
Does not exist
9000141902
Level:
A
Given the function \(f\),
find \(\lim _{x\to \infty }f(x)\).
\[
f(x)=\begin{cases}
x^3+1 & \text{if } x\neq 1,\\
3 & \text{if } x = 1
\end{cases}
\]
\(\infty \)
\(-\infty \)
\(4\)
Does not exist
9000141903
Level:
A
Given the function \(g\),
find \(\lim _{x\to 1^{-}}g(x)\).
\[
g(x)=\begin{cases}
-\frac12(x-1)^2+2 & \text{if } x < 1,\\
\frac2{x^2}+1 & \text{if } x \geq 1
\end{cases}
\]
\(2\)
\(3\)
\(1\)
Does not exist
9000141904
Level:
A
Given the function \(g\),
find \(\lim _{x\to 1^{+}}g(x)\).
\[
g(x)=\begin{cases}
-\frac12(x-1)^2+2 & \text{if } x < 1,\\
\frac2{x^2}+1 & \text{if } x \geq 1
\end{cases}
\]
\(3\)
\(2\)
\(1\)
Does not exist
9000141905
Level:
A
Given the function \(g\) (see the picture),
find \(\lim _{x\to 1}g(x)\).
\[
g(x)=\begin{cases}
-\frac12(x-1)^2+2 & \text{if } x < 1,\\
\frac2{x^2}+1 & \text{if } x \geq 1
\end{cases}
\]
This limit does not exist.
\(3\)
\(2\)
\(1\)
9000141906
Level:
A
Given the function \(g\) (see the picture),
find \(\lim _{x\to \infty }g(x)\).
\[
g(x)=\begin{cases}
-\frac12(x-1)^2+2 & \text{if } x < 1,\\
\frac2{x^2}+1 & \text{if } x \geq 1
\end{cases}
\]
\(1\)
\(0\)
\(\infty \)
\(-\infty \)
This limit does not exist.
9000141907
Level:
A
Given the function \(h\) (see the picture),
find \(\lim _{x\to 1^{-}}h(x)\).
\[
h(x)=\begin{cases}
-\frac1{x-1} & \text{if } x< 1,\\
-(x-1)^2+2 & \text{if } x\geq 1
\end{cases}
\]
\(\infty \)
\(1\)
\(2\)
\(-\infty \)
This limit does not exist.
9000141908
Level:
A
Given the function \(h\),
find \(\lim _{x\to 1^{+}}h(x)\).
\[
h(x)=\begin{cases}
-\frac1{x-1} & \text{if } x< 1,\\
-(x-1)^2+2 & \text{if } x\geq 1
\end{cases}
\]
\(2\)
\(1\)
\(0\)
\(\infty \)
Does not exist