RD Sharma Class 11 Solutions Chapter 29 - Limits (Ex 29.5) Exercise 29.5 - Free PDF
Free PDF download of RD Sharma Class 11 Solutions Chapter 29 - Limits Exercise 29.5 solved by Expert Mathematics Teachers on Vedantu. All Chapter 29 - Limits Ex 29.5 Questions with Solutions for RD Sharma Class 11 Math to help you to revise complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced) and other Engineering entrance exams.
The chapter begins with a section of definitions and basic concepts. An equation in its simplest form is introduced. This equation is presented for a function f(x) for which the derivative is equal to 1. It is also presented in its form which is the standard form of equations in terms of the concept of derivatives.
Here you can learn the mathematical basis of the solution to the limits, ranges, and range limits problems. All the solutions are provided in detail in the pdf document.
The solution to the range and range limits problems requires you to find the range of values to which a function f(x) is confined. This is done by solving the equation f(x) - 1 = 0, which can be written in the form
x2 – 1 = 0,
x = ± 1.
The solutions to the range limit problems are (x – 1) = 0 and (x + 1) = 0.
The solution to the range limits problem is (x – 1) + (x + 1) = 0.
The solution to the range limit problem may be written as (x – 1) + (x + 1) = 0.
Here x is the number that you are going to find the range of and 1 is the number that is contained within the range. The solution to the range limit problem can be written as
(x – 1) + (x + 1) = 0.
To simplify this equation you can multiply the denominator and the numerator by the common factor 3.
This makes the equation
[3(x – 1) + 3(x + 1)] = 0.
x – 3(x + 1) + 3x + 3 = 0.
The x in the equation is positive because the inequality is always written in the form for which x is always greater than or equal to 0.
To simplify the equation, set x + 3 = y, so that the equation becomes
x – 3y + 3 = 0.
Factor out x – 3y, which gives
x – 3y + 3 = 0,
(x – 3y)(x – 3) = 0.
Factoring by grouping the x – 3y and 3 so that you get a product with x2 – 9y2, you get
(x – 3y)(x – 3) = 0,
x2 – 9y2 = 0,
x2 – (x – 3)2 = 0.
Here x – 3 is a factor that is common to the two terms. This is called a quadratic factor and can be factored into
x – (x – 3) = 3.
Because it is a factor of both terms in the factorization, it is a factor that will cancel out of the product. It does not cancel out of the sum because it is not present in the sum.
Now you have an equation of the form y2 – 3y – 9 = 0, or
y2 – 3y + 9 = 0.
Set y = u + v and subtract the equation from itself, which gives
u2 + v2 – 3uv – 9 = 0.
Solve for v and get
v = –(3/2)u – 3/2 or
v = –3u – 9/2
v = –3u + 3 – 3/2.
Add 3 to both sides to get
v = –3u – 9
which is what you wanted.
A limit is a quantity that cannot be expressed as an infinite sum of smaller quantities. Many types of limits can be computed as a finite sum that comes close to the final value as the sum gets larger. For example, here are three summations, each of which is the sum of ten terms:
The first is one hundred thousand terms, the second is one million terms, and the third is one trillion terms. The sum of these three different sommations will be greater than the number of atoms in the known universe.
The third is a summation of about thirty million terms. This sum is only an estimate, but it is a pretty good estimate.
How does this happen? Each term of a sum can be divided by the size of the terms, giving a fraction, and the larger the terms, the smaller the fraction. If a summation of two billion terms is divided by the two billion terms, the resulting fraction is an extremely small number, and if the number is multiplied by itself, the result is still a tiny number. This is called a limit. This means that the term becomes smaller and smaller, approaching zero, but the limit is always greater than zero.
When evaluating the greatest and least limits, you should remember to think about how each number you are using is related to the others. Also, think about how the numbers will fit on the line. Since limits are on a line, if you are looking at an infinite series, think about how it will fit in the space that you are in.
FAQs on RD Sharma Class 11 Solutions Chapter 29 - Limits (Ex 29.5) Exercise 29.5
1. Where can I find Class 11- Limits solutions ?
Class 11 has various chapters in its syllabus, one of them being the chapter on limits. You can find the solutions of this chapter at Vedantu which have been answered by expert teachers because Vedantu aims to be a helping hand to students thereby making way to their desired institutions with detailed notes, syllabus, mock test papers, previous year papers with expert teacher answers and live doubts session. Visit the app or website to know more.
2. Why should Class 11 students follow RD Sharma?
Class 11 students are exposed to an enhanced level of mathematics which might be initially tough for students. Students are advised to solve various types of problems in a concept that gives a better understanding to them. RD Sharma has various questions that have a vast range of questions covered starting from extremely basic to Olympiad level which gives a great exposure to students about the subject.
3. How to solve Math to retain concepts for a longer time?
Mathematics isn’t a subject that can be memorized, whereas it is one such subject that requires conceptual understanding to solve problems and score well in exams. Solving different types of problems in the same concept would benefit students at a later date. Practice daily and for detailed notes, keywords that’ll be handy, students can join us at the Vedantu app for a live doubt clarification session.
4. How to manage time while writing a Mathematics exam?
Many students say that though they went prepared to give their mathematics exam, due to the lengthy and tricky questions, it became difficult. To avoid such experience students are recommended to practice writing faster and memorizing the keywords. Also working out objective type questions beforehand saves time in analyzing and answering them. Lastly, students are advised to divide their time into sections namely A, B, C and D and attempt all the questions without fail.
5. Which textbook/reference book should be used for Class 11 Mathematics?
Mathematics is a very interesting subject that demands time and practice. Students generally panic at the last moment due to a lack of clarity in subject and indefinite sums. For a class 11 student, a few suggested readings could be the NCERT textbook, RS Agarwal and RD Sharma. In addition to this Vedantu also provides students with adequate study material and notes which makes it not only easier but also interesting to the students and helps them grasp and retain the concepts for a longer period. To find notes, visit the Vedantu app or website.
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