Determining Limits Using Algebraic Properties Practice

Sample questions

5 of 8 — sign in to practice the rest with adaptive difficulty and mastery tracking.

1. Sample 1difficulty 1/5

   Evaluate $\displaystyle\lim_{x\to 1}\dfrac{x^2 + 3}{2x - 1}$.

   • A

     $4$

     check_circle
   • B

     $2$

   • C

     $3$

   • D

     $5$

   Why

   Both top and bottom continuous at $x = 1$ (denominator $\ne 0$): $(1 + 3)/(2 - 1) = 4/1 = 4$.

   View full solution & statsarrow_forward

2. Sample 2difficulty 1/5

   Evaluate $\displaystyle\lim_{x \to 3}\left(2x^2 - 5x + 1\right)$.

   • A

     $22$

   • B

     $0$

   • C

     $4$

     check_circle
   • D

     $10$

   Why

   The polynomial is continuous everywhere, so direct substitution works: $2(9) - 5(3) + 1 = 18 - 15 + 1 = 4$.

   View full solution & statsarrow_forward

3. Sample 3difficulty 1/5

   Evaluate $\displaystyle\lim_{x\to\pi/2} \sin x$.

   • A

     $\tfrac{\sqrt 2}{2}$

   • B

     $1$

     check_circle
   • C

     $\tfrac{1}{2}$

   • D

     $0$

   Why

   $\sin$ is continuous everywhere; $\sin(\pi/2) = 1$.

   View full solution & statsarrow_forward

4. Sample 4difficulty 1/5

   Evaluate $\displaystyle\lim_{x\to -1}(x^3 + 2x^2 - 4)$.

   • A

     $-7$

   • B

     $-1$

   • C

     $-3$

     check_circle
   • D

     $0$

   Why

   Polynomial is continuous; substitute: $(-1)^3 + 2(1) - 4 = -1 + 2 - 4 = -3$.

   View full solution & statsarrow_forward

5. Sample 5difficulty 1/5

   Evaluate $\displaystyle\lim_{x\to 7}(\pi)$.

   • A

     Does not exist

   • B

     $\pi$

     check_circle
   • C

     $7$

   • D

     $0$

   Why

   Limit of a constant is the constant itself, regardless of $x \to a$.

   View full solution & statsarrow_forward