• In this chapter, we will develop the concept of a limit by example. Limits and Continuity Basic concepts and formulas based on limit and continuity important for the students of classes 11 and 12. 11. Example Definitions Formulaes. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 $\frac{{{{\rm{x}}^{ - 3}} - {2^{ - 3}}}}{{{\rm{x}} - 2}}$$\left[ {\frac{0}{0}{\rm{form}}} \right]$. Set 2: Multiple-Choice Questions on Limits and Continuity 1. {\rm{x}} + 2}}{{{\rm{x}} + 1}}$ = $\frac{{{\rm{x}} + 2}}{{{\rm{x}} + 1}}$. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{\rm{a}}}{{1 + \sqrt {1 - \frac{{\rm{a}}}{{\rm{x}}}} {\rm{\: }}}}{\rm{\: \: }}$= $\frac{{\rm{a}}}{{1 + \sqrt {1 + 0} }}$ = $\frac{{\rm{a}}}{2}$. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\left( {\frac{1}{{{\rm{x}} - 3}} - \frac{9}{{{{\rm{x}}^2}\left( {{\rm{x}} - 3} \right)}}} \right)$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{{{\rm{x}}^2} - 9}}{{{{\rm{x}}^2}\left( {{\rm{x}} - 3} \right)}}$. NCERT Exemplar Class 11 Maths is very important resource for students preparing for XI Board Examination. In these limits and derivatives Class 11 important questions PDF, we have attached complete concepts of limits and derivatives along with their properties, and formulas are discussed. So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(x) = 3. Or, f(x) = $\frac{{1. • Properties of limits will be established along the way. Limits and Continuity CA CPT. Put x =${\rm{\: }}\frac{{\rm{\pi }}}{2} - {\rm{h}}$. = x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – f(x). Related questions. Continuity of a Function. = secx + $\frac{{{\rm{x}}.2{\rm{sinx}}}}{{{\rm{cosx}}. Introduction to Limits. Variations on the limit theme25 5. = h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |2 + h – 2| = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |h| = 0. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π $\frac{{1 - {\rm{sinx}}/2}}{{{{\left( {{\rm{\pi }} - {\rm{x}}} \right)}^2}}}{\rm{\: }}$= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - \sin \left( {\frac{{{\rm{\pi }} - {\rm{h}}}}{2}} \right)}}{{\frac{{\rm{\pi }}}{2} - \left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}$. = h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {0 + {\rm{h}}} \right|}}{{0 + {\rm{h}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\rm{h}}}{{\rm{h}}}$ = 1. = x $\begin{array}{*{20}{c}}{{\rm{lim}}180}\\\to\end{array}$ 0 = $\frac{{\frac{{\sin \frac{{{\rm{\pi x}}}}{{180}}}}{{\rm{x}}}}}{{\frac{{{\rm{\pi x}}}}{{180}}}}$.$\frac{{\rm{\pi }}}{{180}}$ = 1.$\frac{{\rm{\pi }}}{{180}}$ = $\frac{{\rm{\pi }}}{{180}}$. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a $\frac{{{{\rm{x}}^3} - {{\rm{a}}^3}}}{{{\rm{x}} - {\rm{a}}}}$ = 27. Continuity34 11. Examples of limit computations27 7. Ltd. Trigonometric Equations and General Values. LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (3x2 + 2) = 3*1 + 2 = 5. Limits and Derivatives Class 11 Maths NCERT Solutions were prepared according to CBSE marking … = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - \sin \left( {\frac{{\rm{\pi }}}{2} - \frac{{\rm{h}}}{2}} \right)}}{{\frac{{\rm{\pi }}}{2} - \left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - \cos \frac{{\rm{h}}}{2}}}{{{{\rm{h}}^2}}}$, = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2{{\sin }^2}\frac{{\rm{h}}}{4}}}{{{{\rm{h}}^2}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2{{\left( {\sin \frac{{\rm{h}}}{4}} \right)}^2}}}{{{{\rm{h}}^2}}}{\rm{\: }}$, = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $2{\left( {\frac{{\sin \frac{{\rm{h}}}{4}}}{{\frac{{\rm{h}}}{4}.4}}} \right)^2}$, = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π/2 $\frac{{1 + {\rm{cos}}2{\rm{x}}}}{{{{\left( {{\rm{\pi }} - 2{\rm{x}}} \right)}^2}}}$$\left( {\frac{0}{0}{\rm{form}}} \right)$. Substituting x = 3, we get = 3 + 3 = 6. Solution First note that the function is defined at the given point x = 1 and its value is 5. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. 2.7: Precise Definitions of Limits 2.8: Continuity • The conventional approach to calculus is founded on limits. Evaluate the Given limit: Solution: Given. Again, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{\rm{ax}} + {\rm{b}}}}{{{\rm{x}} + 1}}$. When x = 0, the given function takes the form 0 0. Limits and Continuity Problems and Solutions. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\rm{e}}^{{\rm{sinx}}}} - {\rm{sinx}} - 1}}{{\rm{x}}}$, $\left( {\frac{0}{0}} \right)$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{{\rm{e}}^{{\rm{sinx}}}} - 1}}{{\rm{x}}}} \right)$ – x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{\rm{sinx}}}}{{\rm{x}}}} \right)$ = 1 – 1 = 0. Or, 1 = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{\rm{a}} + \frac{{\rm{b}}}{{\rm{x}}}}}{{1 + \frac{1}{{\rm{x}}}}}$. =  x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left( {{{\left( {1 + {\rm{x}}} \right)}^6} - 1} \right)}}{{{{\left( {1 + {\rm{x}}} \right)}^2} - 1}}$$\left[ {\frac{0}{0}{\rm{form}}} \right]$. All questions and answers from the Past Many Years Question Papers Book of IIT JEE (Advanced) Mathematics Chapter Limit, Continuity & Differentiabilityare provided here for . There may also be additional practice questions. Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Or, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{x}} + 6}}{{{\rm{cx}} - {\rm{d}}}}$. LIMITS AND CONTINUITY WORKSHEET WITH ANSWERS. The concepts in Limits and Derivatives Class 11 NCERT solutions are – (1) Defining derivative of a function, (2) Description of Limits, (3) Limits of trigonometric functions and (4) Derivatives. A limit is a number that a function approaches as the independent variable of the function approaches a given value. = 2cos $\frac{{2{\rm{a}}}}{2}\left( {\sqrt {\rm{a}}  + \sqrt {\rm{a}} } \right).\frac{1}{2}$ = $2\sqrt {\rm{a}} $cosa. = x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (4x2 – 1) = 4*1 – 1 = 3. Revise with Concepts. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{{\rm{e}}^{\rm{x}}} - 1}}{{\rm{x}}}} \right)$ - x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{{\rm{e}}^{ - {\rm{x}}}} - 1}}{{\rm{x}}}} \right)$ + 1 = 1 + 1 + 1 = 3. Related questions. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{{\left( {2 - \frac{1}{{\rm{x}}}} \right)}^6}{{\left( {3 - \frac{1}{{\rm{x}}}} \right)}^4}}}{{{{\left( {2 + \frac{1}{{\rm{x}}}} \right)}^{10}}}}$ = $\frac{{{2^6}{{.3}^4}}}{{{2^{10}}}}$ = $\frac{{{3^4}}}{{{2^4}}}$ = $\frac{{81}}{{16}}$. x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) and x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x) are finite.x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x), For the limit: x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |x| = 0LHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |0 – h| = 0RHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |0 + h| = 0, So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |x| = 0. = x lim → 0 1 − cos9x x2 = x lim → 0 2sin29x 2 x2 = x lim → 0 2(sin 9x 2)2 x2 = x lim → 0 [2(sin 9x 2 9x 2)2(81 4)] = 81 2. = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π/2 $\frac{{{\rm{cosx}}}}{{\frac{{\rm{\pi }}}{2} - {\rm{x}}}}{\rm{\: }}$= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\cos \left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}{{\frac{{\rm{\pi }}}{2} - \left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}$. Students can solve NCERT Class 12 Maths Continuity and Differentiability MCQs Pdf with Answers to know their preparation […] The value L which the function f(x) approaches when x = a is known as the limit of f(x). Combination of these concepts have been widely explained in Class 11 and Class 12. Important formulas of chapter 13 class 11 and important formulas necessary for the students of class 12 chapter 5 Or, f(-2) = $\frac{{\left( { - 2 + 2} \right)}}{{2 + 1}}$ = 0. Or, 1 = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{\rm{ax}} + {\rm{b}}}}{{{\rm{x}} + 1}}$. Then find the limit of the function at x = 1. =  x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{sin}}{{\rm{x}}^0}}}{{\rm{x}}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sin \frac{{{\rm{\pi x}}}}{{180}}}}{{\rm{x}}}$. Informal de nition of limits21 2. Learn Videos. = x l i m → 0 + f (x) = x l i m → 0 + (x + 2) = 0 + 2 = 0, For x < 0, f (x) = 4x + 2. Chapter 01: Continuity × Subtopics 1.0 Introduction (Revision) 1.1 Continuity of a Function at a Point 1.2 Discontinuity of a Function 1.3 Types of Discontinuity 1.4 Algebra of Continuous Functions 1.5 Continuity in an Interval 1.6 Continuity in the Domain of the Function 1.7 Continuity of some Standard Functions NCERT Solutions For Class 11. • We will use limits to analyze asymptotic behaviors of … Problem solving - use acquired knowledge to solve one-sided limits and continuity practice problems Knowledge application - use your knowledge to answer questions about one-sided limits and continuity = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π $\frac{{1 - {\rm{sinx}}/2}}{{{{\left( {{\rm{\pi }} - {\rm{x}}} \right)}^2}}}$$\left( {\frac{0}{0}{\rm{form}}} \right)$. 2 mins read. =x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{{\left( {2{\rm{x}} - 1} \right)}^6}{{\left( {3{\rm{x}} - 1} \right)}^4}}}{{{{\left( {2{\rm{x}} + 1} \right)}^{10}}}}$$\left( {\frac{{\rm{\infty }}}{{\rm{\infty }}}{\rm{form}}} \right)$.