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Calculus 1Calculus 170 views·Updated May 26, 2026·7 pages

Mastering Calculus A: Practice Problems on Limits

user profile picture
Shreya Bedi@shreyabedi_oyaa

Calculus limits are all about understanding what happens to functions... Show more

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of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Evaluating Indeterminate Forms

When you see a limit that gives 00\frac{0}{0} at first, don't panic! This is just an indeterminate form telling you to dig deeper.

For example, in lim⁡x→4x2+x−20x−4\lim_{x \to 4} \frac{x^2+x-20}{x-4}, we get 00\frac{0}{0} initially. The trick is to factor the numerator: (x+5)(x−4)(x+5)(x-4), which lets us cancel the (x−4)(x-4) term. The limit simplifies to x+5x+5 evaluated at x=4x=4, giving us 9.

💡 Quick Tip: When a limit gives you 00\frac{0}{0}, try factoring or using algebraic manipulation to cancel common terms before evaluating.

With difference quotients like lim⁡h→0(−9+h)2−81h\lim_{h \to 0} \frac{(-9+h)^2 - 81}{h}, expand and simplify first: h2−18hh=h−18\frac{h^2-18h}{h} = h-18. When hh approaches 0, this becomes $0-18 = -18$.

2
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Understanding Infinite Limits

Functions can behave dramatically when approaching certain values - shooting up to infinity, diving down to negative infinity, or approaching a specific value.

When evaluating limits with fractions where the denominator approaches zero, the limit often goes to infinity. For example, lim⁡x→56−x(x−5)\lim_{x \to 5} \frac{6-x}{(x-5)} approaches negative infinity because as xx gets closer to 5, the denominator gets extremely small while the numerator approaches 1.

🔍 Remember: When the denominator approaches zero and the numerator doesn't, your limit will head toward infinity (positive or negative depending on the signs).

For rational functions with large inputs, divide both numerator and denominator by the highest power of the variable. In lim⁡t→∞7t+t2t−t2\lim_{t \to \infty} \frac{7t + t^2}{t - t^2}, dividing by t2t^2 simplifies it to lim⁡t→∞7/t+11/t−1\lim_{t \to \infty} \frac{7/t + 1}{1/t - 1}, which equals -1 as tt approaches infinity.

3
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Limits and Continuity

Evaluating complex limits becomes easier when you break them down using limit properties. For example, lim⁡x→2(5x2−3x−3)\lim_{x \to 2} (5x^2 - 3x - 3) can be calculated by evaluating each term separately and then combining: $5(2)^2 - 3(2) - 3 = 20 - 6 - 3 = 11$.

Vertical asymptotes (VA) occur when a function's denominator equals zero, making the function undefined. Horizontal asymptotes (HA) appear when x approaches infinity and the function approaches a constant value.

📊 Visual Aid: Think of asymptotes as "barriers" the function gets extremely close to but never touches or crosses.

Average velocity is calculated as S(t2)−S(t1)t2−t1\frac{S(t_2) - S(t_1)}{t_2 - t_1}, which is the change in position divided by the change in time. For instance, if a position changes from 78.7 to 20.1 in 2 seconds, the average velocity is 78.7−20.12=29.3\frac{78.7 - 20.1}{2} = 29.3 ft/s.

4
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Special Limit Cases

When evaluating limits involving square roots, rationalizing can help. For expressions like lim⁡h→064+h−8h\lim_{h \to 0} \frac{\sqrt{64 + h} - 8}{h}, multiplying both numerator and denominator by the conjugate helps eliminate the indeterminate form.

For functions like f(x)=3x2−x4f(x) = 3x^2 - x^4, understanding end behavior is crucial. As xx approaches infinity, the −x4-x^4 term dominates, making the limit approach negative infinity. Writing it as x2(3−x2)x^2(3 - x^2) helps visualize this behavior.

🧠 Think About It: The highest-power term usually determines how a function behaves as x approaches infinity.

When analyzing limits of piecewise functions, you need to evaluate each piece separately. For a function defined differently for x<0x < 0, $0 \leq x \leq 1,and, and x > 1,examinethelimitfrombothdirectionswhenapproachingboundarypointslike, examine the limit from both directions when approaching boundary points like x = 0or or x = 1$.

5
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Limit Applications and Properties

When dealing with combined functions, apply limit properties. For instance, lim⁡x→0x+3x24x−1/x\lim_{x \to 0} \frac{\sqrt{x} + 3x^2}{4x - 1/x} can be simplified by analyzing which terms dominate as x approaches zero.

The squeeze theorem is incredibly useful when functions are bounded. If $3x - 1 \leq f(x) \leq x^2 - 3x + 8andbothboundsequal8when and both bounds equal 8 when x = 3,then, then \lim_{x \to 3} f(x) = 8$.

🔑 Key Insight: Limit properties let you break complex expressions into manageable pieces. Remember that lim⁡x→a[f(x)+g(x)]=lim⁡x→af(x)+lim⁡x→ag(x)\lim_{x \to a}[f(x) + g(x)] = \lim_{x \to a}f(x) + \lim_{x \to a}g(x).

When analyzing function behavior near vertical asymptotes, check the limit from both directions. For lim⁡x→5x2−5xx2−10x+25\lim_{x \to 5} \frac{x^2 - 5x}{x^2 - 10x + 25}, simplifying to xx−5\frac{x}{x-5} shows that the limit approaches +∞+\infty from the right and −∞-\infty from the left.

6
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Working with Rational Functions

When evaluating limits of rational functions as x approaches infinity, focus on the terms with the highest powers in both numerator and denominator. This determines the end behavior of the function.

For complex fractions like lim⁡x→∞(5x2+1)(x−7)2(x2+x)\lim_{x \to \infty} \frac{(5x^2+1)}{(x-7)^2(x^2+x)}, divide every term by the highest power of x in this case $x^4$ to simplify the expression.

⚡ Power Tip: When x approaches infinity, terms with lower powers become negligible compared to those with higher powers.

This approach transforms complicated rational functions into manageable expressions where most terms approach zero, leaving only the ratio of the coefficients of the highest-power terms.

7
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

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Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

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Calculus 1Calculus 170 views·Updated May 26, 2026·7 pages

Mastering Calculus A: Practice Problems on Limits

user profile picture
Shreya Bedi@shreyabedi_oyaa

Calculus limits are all about understanding what happens to functions when they approach certain values. Think of limits as a mathematical zoom lens, letting you see exactly where a function is heading, even at tricky points where the function might... Show more

1
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

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  • Join milions of students

Evaluating Indeterminate Forms

When you see a limit that gives 00\frac{0}{0} at first, don't panic! This is just an indeterminate form telling you to dig deeper.

For example, in lim⁡x→4x2+x−20x−4\lim_{x \to 4} \frac{x^2+x-20}{x-4}, we get 00\frac{0}{0} initially. The trick is to factor the numerator: (x+5)(x−4)(x+5)(x-4), which lets us cancel the (x−4)(x-4) term. The limit simplifies to x+5x+5 evaluated at x=4x=4, giving us 9.

💡 Quick Tip: When a limit gives you 00\frac{0}{0}, try factoring or using algebraic manipulation to cancel common terms before evaluating.

With difference quotients like lim⁡h→0(−9+h)2−81h\lim_{h \to 0} \frac{(-9+h)^2 - 81}{h}, expand and simplify first: h2−18hh=h−18\frac{h^2-18h}{h} = h-18. When hh approaches 0, this becomes $0-18 = -18$.

2
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

Understanding Infinite Limits

Functions can behave dramatically when approaching certain values - shooting up to infinity, diving down to negative infinity, or approaching a specific value.

When evaluating limits with fractions where the denominator approaches zero, the limit often goes to infinity. For example, lim⁡x→56−x(x−5)\lim_{x \to 5} \frac{6-x}{(x-5)} approaches negative infinity because as xx gets closer to 5, the denominator gets extremely small while the numerator approaches 1.

🔍 Remember: When the denominator approaches zero and the numerator doesn't, your limit will head toward infinity (positive or negative depending on the signs).

For rational functions with large inputs, divide both numerator and denominator by the highest power of the variable. In lim⁡t→∞7t+t2t−t2\lim_{t \to \infty} \frac{7t + t^2}{t - t^2}, dividing by t2t^2 simplifies it to lim⁡t→∞7/t+11/t−1\lim_{t \to \infty} \frac{7/t + 1}{1/t - 1}, which equals -1 as tt approaches infinity.

3
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

Limits and Continuity

Evaluating complex limits becomes easier when you break them down using limit properties. For example, lim⁡x→2(5x2−3x−3)\lim_{x \to 2} (5x^2 - 3x - 3) can be calculated by evaluating each term separately and then combining: $5(2)^2 - 3(2) - 3 = 20 - 6 - 3 = 11$.

Vertical asymptotes (VA) occur when a function's denominator equals zero, making the function undefined. Horizontal asymptotes (HA) appear when x approaches infinity and the function approaches a constant value.

📊 Visual Aid: Think of asymptotes as "barriers" the function gets extremely close to but never touches or crosses.

Average velocity is calculated as S(t2)−S(t1)t2−t1\frac{S(t_2) - S(t_1)}{t_2 - t_1}, which is the change in position divided by the change in time. For instance, if a position changes from 78.7 to 20.1 in 2 seconds, the average velocity is 78.7−20.12=29.3\frac{78.7 - 20.1}{2} = 29.3 ft/s.

4
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

Special Limit Cases

When evaluating limits involving square roots, rationalizing can help. For expressions like lim⁡h→064+h−8h\lim_{h \to 0} \frac{\sqrt{64 + h} - 8}{h}, multiplying both numerator and denominator by the conjugate helps eliminate the indeterminate form.

For functions like f(x)=3x2−x4f(x) = 3x^2 - x^4, understanding end behavior is crucial. As xx approaches infinity, the −x4-x^4 term dominates, making the limit approach negative infinity. Writing it as x2(3−x2)x^2(3 - x^2) helps visualize this behavior.

🧠 Think About It: The highest-power term usually determines how a function behaves as x approaches infinity.

When analyzing limits of piecewise functions, you need to evaluate each piece separately. For a function defined differently for x<0x < 0, $0 \leq x \leq 1,and, and x > 1,examinethelimitfrombothdirectionswhenapproachingboundarypointslike, examine the limit from both directions when approaching boundary points like x = 0or or x = 1$.

5
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

Limit Applications and Properties

When dealing with combined functions, apply limit properties. For instance, lim⁡x→0x+3x24x−1/x\lim_{x \to 0} \frac{\sqrt{x} + 3x^2}{4x - 1/x} can be simplified by analyzing which terms dominate as x approaches zero.

The squeeze theorem is incredibly useful when functions are bounded. If $3x - 1 \leq f(x) \leq x^2 - 3x + 8andbothboundsequal8when and both bounds equal 8 when x = 3,then, then \lim_{x \to 3} f(x) = 8$.

🔑 Key Insight: Limit properties let you break complex expressions into manageable pieces. Remember that lim⁡x→a[f(x)+g(x)]=lim⁡x→af(x)+lim⁡x→ag(x)\lim_{x \to a}[f(x) + g(x)] = \lim_{x \to a}f(x) + \lim_{x \to a}g(x).

When analyzing function behavior near vertical asymptotes, check the limit from both directions. For lim⁡x→5x2−5xx2−10x+25\lim_{x \to 5} \frac{x^2 - 5x}{x^2 - 10x + 25}, simplifying to xx−5\frac{x}{x-5} shows that the limit approaches +∞+\infty from the right and −∞-\infty from the left.

6
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

Working with Rational Functions

When evaluating limits of rational functions as x approaches infinity, focus on the terms with the highest powers in both numerator and denominator. This determines the end behavior of the function.

For complex fractions like lim⁡x→∞(5x2+1)(x−7)2(x2+x)\lim_{x \to \infty} \frac{(5x^2+1)}{(x-7)^2(x^2+x)}, divide every term by the highest power of x in this case $x^4$ to simplify the expression.

⚡ Power Tip: When x approaches infinity, terms with lower powers become negligible compared to those with higher powers.

This approach transforms complicated rational functions into manageable expressions where most terms approach zero, leaving only the ratio of the coefficients of the highest-power terms.

7
of 7
Calculus A: Limits Practice Problems ①g(x)=$x+x-20$ $|x-4|$ lim $g(x)$= $16+4-20$ = $\frac{0}{0}$ $x\rightarrow4$ $\frac{14-41}{14-41}$

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
  • Join milions of students

We thought you’d never ask...

What is the Knowunity AI companion?

Our AI companion is specifically built for the needs of students. Based on the millions of content pieces we have on the platform we can provide truly meaningful and relevant answers to students. But its not only about answers, the companion is even more about guiding students through their daily learning challenges, with personalised study plans, quizzes or content pieces in the chat and 100% personalisation based on the students skills and developments.

Where can I download the Knowunity app?

You can download the app in the Google Play Store and in the Apple App Store.

Is Knowunity really free of charge?

That's right! Enjoy free access to study content, connect with fellow students, and get instant help – all at your fingertips.

Most popular content in Calculus 1

7

Most popular content

9

Can't find what you're looking for? Explore other subjects.

Students love us — and so will you.

4.6/5App Store
4.7/5Google Play

The app is very easy to use and well designed. I have found everything I was looking for so far and have been able to learn a lot from the presentations! I will definitely use the app for a class assignment! And of course it also helps a lot as an inspiration.

Stefan SiOS user

This app is really great. There are so many study notes and help [...]. My problem subject is French, for example, and the app has so many options for help. Thanks to this app, I have improved my French. I would recommend it to anyone.

Samantha KlichAndroid user

Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

AnnaiOS user