11th Class Mathematics Limits and Derivatives Notes - Mathematics Olympiads - Limits and Derivatives

 

Limits and Derivatives

 

Key Points to Remember

 

A number, (\[\ell \] is said to be the limit of the function \[\text{y=f}(x)\] at\[x=a\], then \[\exists \] a positive number, \[\in \,\,>\,\,0\] corresponding to the small positive number \[\delta \,>\,\,0\] such that

\[\left| \text{f(x)-}\left. \ell  \right| \right.\,<\,\varepsilon ,\] provided \[\left| x-\left. a \right| \right.\,<\,\delta \]

Evidently,

\[\underset{x\to \,a}{\mathop{\lim }}\,\,\,f(x)=\ell \]

We have to learn how to evaluate the limit of the function \[y=f(x)\]

Only three type of function can be evaluate the limit of the function.

1. Algebraic function
2. Trigonometric function
3. Logarithmic and exponential function

Now, basically, there are three method to evaluate the limit of the function.

 

• By Formula Method

i.e.        \[\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n.{{a}^{n-1}}\]

 

• Factorisation Method: In this matter numerator and denominator are factorised. The common factors are cancelled and the rest is the result.
• Rationalisation Method: Rationalisation is followed when we have fractional overs \[\left( like\frac{1}{2},\frac{1}{3}etc. \right)\] on expressions in numerator or denominator in both. After rationalisation the terms are factorised which on cancellation gives the result.

 

1. Q. \[\underset{\mathbf{x3}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-27}}{\mathbf{x-3}}\mathbf{=}\underset{\mathbf{x3}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-(3}{{\mathbf{)}}^{\mathbf{3}}}}{\mathbf{x-3}}\mathbf{=3(3}{{\mathbf{)}}^{\mathbf{3-1}}}\]

 

\[={{3.3}^{2}}=27\]

 

1. \[\underset{\mathbf{x1}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{15}}}\mathbf{-1}}{{{\mathbf{x}}^{\mathbf{10}}}\mathbf{-1}}\]

 

Dividing numerator and denomiator by (x -1), when

 

\[=\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{15}}-{{1}^{15}}}{\frac{x-1}{\frac{{{x}^{^{10}}}-{{(1)}^{10}}}{x-1}}}=\frac{15.{{(1)}^{15-1}}}{10.{{(1)}^{10-1}}}=\frac{15}{10}=\frac{3}{2}\]

 

 

1. \[\underset{\mathbf{x0}}{\mathop{\mathbf{lim}}}\,\frac{\sqrt{\mathbf{1+x-1}}}{\mathbf{x}}\mathbf{=}\underset{\mathbf{x0}}{\mathop{\mathbf{lim}}}\,\frac{\mathbf{(}\sqrt{\mathbf{1+x-1)}}}{\mathbf{x}}\mathbf{\times }\frac{\sqrt{\mathbf{1+x}}\mathbf{+1}}{\sqrt{\mathbf{1+x}}\mathbf{+1}}\]

 

            \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1+x-1}{x(\sqrt{1+x+1)}}[\therefore \,\,{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)]=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{(\sqrt{1+x+1)}};\]

applying limit, \[x\to 0\],we have \[=\frac{1}{1+1}=\frac{1}{2}\]

 

1. \[\underset{\mathbf{x2}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+4}}{\mathbf{x-2}}\mathbf{=}\underset{\mathbf{x2}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}{{\mathbf{2}}^{\mathbf{2}}}}{\mathbf{x-2}}\]

 

                        \[\underset{x\to 2}{\mathop{\lim }}\,\frac{(x+2)(x-2)}{x-2}=2+2=4\]

 

Q,        \[\underset{\mathbf{x3}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{4}}}\mathbf{-81}}{\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-5x-3}}\mathbf{=}\underset{\mathbf{x3}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{4}}}\mathbf{-(3}{{\mathbf{)}}^{\mathbf{4}}}}{\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-6x+x-3}}\]

 

\[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{({{x}^{2}})}^{2}}-{{({{3}^{2}})}^{2}}}{2x(x-3)+1(x-3)}=\underset{x\to 3}{\mathop{\lim }}\,\frac{({{x}^{2}}-9)-({{x}^{2}}+9)}{(x-3)(2x+1)}=\underset{x\to 3}{\mathop{\lim }}\,\frac{(x+3)(x-3)({{x}^{2}}+9)}{(x-3)(2x+1)}\]

 

Applying limit as \[x\to 3\], we have

 

\[=\frac{(3+3)({{3}^{2}}+9)}{(2\times 3+1)}=\frac{6\times 18}{7}=\frac{108}{7}\]

 

• Properties of Limits

 

If \[\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)=}\ell \]and \[\underset{x\to a}{\mathop{\lim }}\,\,\,g\text{(x)=m}\] then the following results are true:

 

(a)        \[\underset{x\to a}{\mathop{\lim }}\,\,\left[ f(x)\pm \,g\text{(x)} \right]\text{=}\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)}\pm \underset{x\to a}{\mathop{\lim }}\,\,\,g(x)=\ell +m\]

 

(b)        \[\underset{x\to a}{\mathop{\lim }}\,\,\left\{ k.\text{f(x)} \right\}\text{=k}\text{.}\,\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)}\text{.}\,=k.\ell .\]

 

(c)        \[\underset{x\to a}{\mathop{\lim }}\,\,\left\{ k.\text{f(x)} \right\}\text{=k}\text{.}\,\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)}\text{.}\,\underset{x\to a}{\mathop{\lim }}\,\,\,g(x)=\ell +m.\]

 

            (d)        \[\underset{x\to a}{\mathop{\lim }}\,\,\left\{ \frac{\text{f(x)}}{g(x)} \right\}\text{=}\frac{\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)}}{\underset{x\to a}{\mathop{\lim }}\,\,\,g\text{(x)}}=\frac{\ell }{m}[i\text{f}\,\,\text{m}\ne \text{0 }\!\!]\!\!\text{ }\]

 

            (e)        \[\underset{x\to a}{\mathop{\lim }}\,\,\text{f(x)=+}\infty \]or\[\text{-}\,\infty \], then \[\underset{x\to a}{\mathop{\lim }}\,\,\frac{1}{\text{f(x)}}\text{=0}\]

 

            (f)         \[\underset{x\to a}{\mathop{\lim }}\,\,\,\log \left\{ g(x) \right\}\,=\log \left[ \underset{x\to a}{\mathop{\lim \,g(x)}}\, \right]=\log \,m,\,m>0\frac{1}{\text{f(x)}}\text{=0}\]

 

            (g)        \[\underset{x\to a}{\mathop{\lim }}\,\,\,{{\left[ f(x) \right]}^{g(x)}}\,={{\{\underset{x\to a}{\mathop{\lim }}\,\,\text{f(x) }\!\!\}\!\!\text{ }}^{\underset{x\to a}{\mathop{\lim \,\,\,g(x)}}\,}}={{\ell }^{m}}\]

 

• Remember these results

 

1. \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\]
2. \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})\]
3. \[{{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}+ab+{{b}^{2}})\]
4. \[{{a}^{4}}-{{b}^{4}}={{({{a}^{2}})}^{2}}-{{({{b}^{2}})}^{2}})=({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}})-(a+b)(a-b)({{a}^{2}}+{{b}^{2}})\]

 

5. \[\sum{n=1+2+3+4....n=\frac{n(n+1)}{2}}\]

 

6. \[\sum{{{n}^{2}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}....{{n}^{2}}=\frac{n(n+1)(2n+1)}{6}}\]

 

7. \[{{\sum{{{n}^{3}}={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}....{{n}^{3}}=\left\{ \frac{n(n+1)}{2} \right\}}}^{2}}\]

 

8. \[\log (1+x)=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+.....\infty \]

 

9. \[\log (1-x)=-x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+.....to\,\infty \]

 

10. \[{{e}^{x}}=1+x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+\frac{{{x}^{4}}}{4}+.....to\infty \]

 

11. \[{{e}^{-x}}=1-x+\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}+\frac{{{x}^{4}}}{4}-.....to\infty \]

 

12. \[\frac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+.....to\,\infty ,\] when x<1

 

13.  \[{{a}^{x}}=1+x\log a+\frac{{{(x\log a)}^{2}}}{2}+.....to\infty \]

 

14. \[\sin x=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}......to\infty \]

 

 

15. \[\cos x=1-\frac{{{x}^{2}}}{2}+\frac{{{x}^{4}}}{4}-\frac{{{x}^{6}}}{6}+......to\infty \]

 

16. \[\tan x=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}+.....to\infty \]

 

17. \[{{\tan }^{-1}}x=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{{{x}^{7}}}{7}+.....0,\frac{\pi }{x}\le x\le \frac{\pi }{4}\] 

• Evaluation of limit of Trigonometrically Functions

Some basic formula,

 

1. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \,x}{x}=1\]
2. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan \,x}{x}=1\]
3. \[\underset{x\to 0}{\mathop{\lim }}\,\cos \,x=1\]
4. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}x}{x}=1\]
5. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{ta{{m}^{-1}}x}{x}=1\]

 

• Some useful results of evaluation of limit of logarthmic and expontential function

 

• \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (1+x)}{x}=1\]

 

• \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{x}}-1}{x}={{\log }_{e}}a\]

 

• \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-1}{x}=1\]

 

• \[\underset{x\to 0}{\mathop{\lim }}\,{{(1+x)}^{\frac{1}{x}}}=e\]

 

 

 

• Limit in the form \[{{1}^{\infty }}\]or \[0{}^\circ \]

 

\[\underset{x\to \infty }{\mathop{\lim }}\,{{\left\{ \text{f(x)} \right\}}^{g(x)}}=\underset{x\to \infty }{\mathop{\lim }}\,\,\,{{e}^{\left\{ \text{f(x)}-1 \right\}.g(x)}}\]

 

e.g.       \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{\left( 1+\frac{1}{x} \right)}^{x\to g(x)}}}{f(x)}=\frac{1}{x}=0\]

 

\[{{(1+0)}^{\infty }}={{1}^{\infty }}=\underset{x\to \infty }{\mathop{\lim }}\,\,\,e\left\{ f(x)-1) \right\}.g(x)=\underset{x\to \infty }{\mathop{\lim }}\,\,\,{{e}^{\left( 1+\frac{1}{x}-1 \right).x}}=\underset{x\to \infty }{\mathop{\lim }}\,\,\,{{e}^{x\times \frac{1}{x}}}={{e}^{1}}=e\]

 

1. \[\underset{\mathbf{n\text{¥}}}{\mathop{\mathbf{lim}}}\,\,\,\mathbf{n(}\sqrt{{{\mathbf{n}}^{\mathbf{2}}}\mathbf{+4}}\mathbf{-n}\]

           

                        \[\underset{n\to \infty }{\mathop{\lim }}\,\,\,n\frac{(\sqrt{{{n}^{2}}+4}-n}{\sqrt{{{n}^{2}}+4}+n}\times (\sqrt{{{n}^{2}}+4}+n)\]                                              \[\left[ \because (a-b)(a+b)={{a}^{2}}-{{b}^{2}} \right]\]  

           

                       

\[\underset{n\to \infty }{\mathop{\lim }}\,\,\,n\frac{(\sqrt{{{n}^{2}}+4}-{{n}^{2}}}{\sqrt{{{n}^{2}}+4}+n}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{4}{\sqrt{\frac{{{n}^{2}}+4}{{{n}^{2}}}}+\frac{n}{n}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{4}{\sqrt{1+\frac{4}{{{n}^{2}}}}+1}\]

                        Applying limit, \[n\to \infty ,\]we have

                       

\[=\frac{4}{\sqrt{1+0}+1}=\frac{4}{2}=2.\left[ \because \frac{1}{\infty }=0 \right]\]

            \[=\frac{4}{\sqrt{1+0}+1}=\frac{4}{2}=2\]

 

 

1. Evaluate \[\underset{\mathbf{x0}}{\mathop{\mathbf{lim}}}\,\frac{\mathbf{2si}{{\mathbf{n}}^{\mathbf{-1}}}\mathbf{x}}{\mathbf{3x}}\]

Sol:      Let \[{{\sin }^{-1}}x=\theta \Rightarrow \sin \theta =x.\] Now, \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2.{{\sin }^{-1}}x}{3x}\]                     \[x\to 0\]

                                                                                                \[\sin \theta \to 0\]

                                                                                                \[\theta =0\]

                        \[\underset{x\to 0}{\mathop{\lim }}\,\frac{2\theta }{3.\sin \theta }\]

 

                        \[\underset{\sin \theta \to 0}{\mathop{\lim }}\,\frac{2}{3\left( \frac{\sin \theta }{\theta } \right)}=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{\left( \frac{2}{3} \right)}{\frac{\sin \theta }{\theta }}=\frac{\frac{2}{3}}{1}=\left( \frac{2}{3} \right)\]

 

 

1. Evaluate \[\underset{\mathbf{x\pi }}{\mathop{\mathbf{lim}}}\,\left( \frac{\mathbf{1+co}{{\mathbf{s}}^{\mathbf{3}}}\mathbf{x}}{\mathbf{ta}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{x}} \right)\]

           

{we know that \[\cos \,\,3x=4{{\cos }^{3}}x-3\cos x\]                  \[\Rightarrow \]   \[4{{\cos }^{3}}x=\cos 3x+3\cos x\]

and \[\sin 3x=3\sin x-4{{\sin }^{3}}x\}\]

                        \[=\underset{x\to \pi }{\mathop{\lim }}\,\left( \frac{1}{{{\tan }^{2}}x}+\frac{{{\cos }^{3}}x}{{{\tan }^{2}}x} \right)=\underset{x\to \pi }{\mathop{\lim }}\,\left( {{\cot }^{2}}x+\frac{{{\cos }^{3}}x}{\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}} \right)\]

                       

\[\underset{x\to \pi }{\mathop{\lim }}\,\left( {{\cot }^{2}}x \right)+{{\cot }^{2}}x.{{\cos }^{3}}x\]

 

                        \[\underset{x\to \pi }{\mathop{\lim }}\,\left( \frac{{{1}^{3}}+{{\cos }^{3}}x}{{{\tan }^{2}}x} \right)\left[ \therefore {{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}}) \right]\]

 

                        \[\underset{x\to \pi }{\mathop{\lim }}\,\frac{(1+\cos x)(1-\cos x+{{\cos }^{2}}x}{({{\tan }^{2}}x)}=\underset{x\to \pi }{\mathop{\lim }}\,\frac{(1+\cos x)(1-\cos x+{{\cos }^{2}}x)}{\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}\]

 

                        \[\underset{x\to \pi }{\mathop{\lim }}\,\frac{{{\cos }^{2}}x(1+\cos x)(1-\cos x+{{\cos }^{2}}x)}{({{1}^{2}}-{{\cos }^{2}}x)}\]

 

                        \[\underset{x\to \pi }{\mathop{\lim }}\,\frac{{{\cos }^{2}}x(1+\cos x)(1-\cos x+{{\cos }^{2}}x)}{(1-\cos x)(1-cosx)}=\frac{{{(-1)}^{2}}(1+1+1)}{1-(-1)}\frac{3}{2}\]

           

                        \[[\cos \,\,\pi =-1]\]