Evaluating Limits — Practice Problems and Worked Solutions

The formulas you need here

26 worked problems

Every step is generated and checked by a computer algebra system.

1. 1.

   Evaluate the limit.

   $$\underset{x\to 1}{\lim }\frac{x^2+3x}{x+2}$$

   Answer: $\frac{4}{3}$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 1}{\lim }\frac{x^2+3x}{x+2}=\frac{4}{3}$$
2. 2.

   Evaluate the limit.

   $$\underset{x\to 2}{\lim }\frac{2x+1}{x+3}$$

   Answer: $1$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 2}{\lim }\frac{2x+1}{x+3}=1$$
3. 3.

   Evaluate the limit.

   $$\underset{x\to 3}{\lim }\frac{x^2-4}{x+1}$$

   Answer: $\frac{5}{4}$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 3}{\lim }\frac{x^2-4}{x+1}=\frac{5}{4}$$
4. 4.

   Evaluate the limit.

   $$\underset{x\to 2}{\lim }\frac{x^3-x}{x^2+1}$$

   Answer: $\frac{6}{5}$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 2}{\lim }\frac{x^3-x}{x^2+1}=\frac{6}{5}$$
5. 5.

   Evaluate the limit.

   $$\underset{x\to 1}{\lim }\frac{3x+2}{x+5}$$

   Answer: $\frac{5}{6}$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 1}{\lim }\frac{3x+2}{x+5}=\frac{5}{6}$$
6. 6.

   Evaluate the limit.

   $$\underset{x\to 2}{\lim }\frac{x^2+1}{x+4}$$

   Answer: $\frac{5}{6}$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 2}{\lim }\frac{x^2+1}{x+4}=\frac{5}{6}$$
7. 7.

   Evaluate the limit.

   $$\underset{x\to 3}{\lim }\frac{4x-1}{x+2}$$

   Answer: $\frac{11}{5}$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 3}{\lim }\frac{4x-1}{x+2}=\frac{11}{5}$$
8. 8.

   Evaluate the limit.

   $$\underset{x\to 2}{\lim }\frac{x^2-9}{x+5}$$

   Answer: $-\frac{5}{7}$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 2}{\lim }\frac{x^2-9}{x+5}=-\frac{5}{7}$$
9. 9.

   Evaluate the limit.

   $$\underset{x\to 1}{\lim }\frac{x^3+2}{x^2+3}$$

   Answer: $\frac{3}{4}$

   Step-by-step solution

   1. $$\text{The denominator is nonzero here, so substitute directly.}$$
   2. $$\underset{x\to 1}{\lim }\frac{x^3+2}{x^2+3}=\frac{3}{4}$$
10. 10.

    Evaluate the limit.

    $$\underset{x\to 2}{\lim }\frac{5x+3}{2x+1}$$

    Answer: $\frac{13}{5}$

    Step-by-step solution

    1. $$\text{The denominator is nonzero here, so substitute directly.}$$
    2. $$\underset{x\to 2}{\lim }\frac{5x+3}{2x+1}=\frac{13}{5}$$
11. 11.

    Evaluate the limit.

    $$\underset{x\to 3}{\lim }\frac{x^2+5x}{x+6}$$

    Answer: $\frac{8}{3}$

    Step-by-step solution

    1. $$\text{The denominator is nonzero here, so substitute directly.}$$
    2. $$\underset{x\to 3}{\lim }\frac{x^2+5x}{x+6}=\frac{8}{3}$$
12. 12.

    Evaluate the limit.

    $$\underset{x\to 3}{\lim }\frac{2x^2-x}{x^2+2}$$

    Answer: $\frac{15}{11}$

    Step-by-step solution

    1. $$\text{The denominator is nonzero here, so substitute directly.}$$
    2. $$\underset{x\to 3}{\lim }\frac{2x^2-x}{x^2+2}=\frac{15}{11}$$
13. 13.

    Evaluate the limit.

    $$\underset{x\to 2}{\lim }\frac{x^3-4x}{x+7}$$

    Answer: $0$

    Step-by-step solution

    1. $$\text{The denominator is nonzero here, so substitute directly.}$$
    2. $$\underset{x\to 2}{\lim }\frac{x^3-4x}{x+7}=0$$
14. 14.

    Evaluate the limit.

    $$\underset{x\to 1}{\lim }\frac{3x^2+x-2}{2x^2+5}$$

    Answer: $\frac{2}{7}$

    Step-by-step solution

    1. $$\text{The denominator is nonzero here, so substitute directly.}$$
    2. $$\underset{x\to 1}{\lim }\frac{3x^2+x-2}{2x^2+5}=\frac{2}{7}$$
15. 15.

    Evaluate the limit.

    $$\underset{x\to 2}{\lim }\frac{x^2-4}{x-2}$$

    Answer: $4$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$$
    3. $$\underset{x\to 2}{\lim }(x+2)=4$$
16. 16.

    Evaluate the limit.

    $$\underset{x\to 3}{\lim }\frac{x^2-9}{x-3}$$

    Answer: $6$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3$$
    3. $$\underset{x\to 3}{\lim }(x+3)=6$$
17. 17.

    Evaluate the limit.

    $$\underset{x\to 5}{\lim }\frac{x^2-25}{x-5}$$

    Answer: $10$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-25}{x-5}=\frac{(x-5)(x+5)}{x-5}=x+5$$
    3. $$\underset{x\to 5}{\lim }(x+5)=10$$
18. 18.

    Evaluate the limit.

    $$\underset{x\to 1}{\lim }\frac{x^2-1}{x-1}$$

    Answer: $2$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1$$
    3. $$\underset{x\to 1}{\lim }(x+1)=2$$
19. 19.

    Evaluate the limit.

    $$\underset{x\to 4}{\lim }\frac{x^2-16}{x-4}$$

    Answer: $8$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}=x+4$$
    3. $$\underset{x\to 4}{\lim }(x+4)=8$$
20. 20.

    Evaluate the limit.

    $$\underset{x\to 6}{\lim }\frac{x^2-36}{x-6}$$

    Answer: $12$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-36}{x-6}=\frac{(x-6)(x+6)}{x-6}=x+6$$
    3. $$\underset{x\to 6}{\lim }(x+6)=12$$
21. 21.

    Evaluate the limit.

    $$\underset{x\to 7}{\lim }\frac{x^2-49}{x-7}$$

    Answer: $14$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-49}{x-7}=\frac{(x-7)(x+7)}{x-7}=x+7$$
    3. $$\underset{x\to 7}{\lim }(x+7)=14$$
22. 22.

    Evaluate the limit.

    $$\underset{x\to 8}{\lim }\frac{x^2-64}{x-8}$$

    Answer: $16$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-64}{x-8}=\frac{(x-8)(x+8)}{x-8}=x+8$$
    3. $$\underset{x\to 8}{\lim }(x+8)=16$$
23. 23.

    Evaluate the limit.

    $$\underset{x\to 9}{\lim }\frac{x^2-81}{x-9}$$

    Answer: $18$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-81}{x-9}=\frac{(x-9)(x+9)}{x-9}=x+9$$
    3. $$\underset{x\to 9}{\lim }(x+9)=18$$
24. 24.

    Evaluate the limit.

    $$\underset{x\to 10}{\lim }\frac{x^2-100}{x-10}$$

    Answer: $20$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-100}{x-10}=\frac{(x-10)(x+10)}{x-10}=x+10$$
    3. $$\underset{x\to 10}{\lim }(x+10)=20$$
25. 25.

    Evaluate the limit.

    $$\underset{x\to 11}{\lim }\frac{x^2-121}{x-11}$$

    Answer: $22$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-121}{x-11}=\frac{(x-11)(x+11)}{x-11}=x+11$$
    3. $$\underset{x\to 11}{\lim }(x+11)=22$$
26. 26.

    Evaluate the limit.

    $$\underset{x\to 12}{\lim }\frac{x^2-144}{x-12}$$

    Answer: $24$

    Step-by-step solution

    1. $$\text{Direct substitution gives }\frac{0}{0}\text{. Factor and cancel.}$$
    2. $$\frac{x^2-144}{x-12}=\frac{(x-12)(x+12)}{x-12}=x+12$$
    3. $$\underset{x\to 12}{\lim }(x+12)=24$$

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