Limits and Derivatives - Solutions

 CBSE Class–11 Mathematics

NCERT Solutions

Chapter - 13 Limits and Derivative

Exercise 13.1

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Evaluate the following limits in Exercises 1 to 22.

1. 

Ans.  3 + 3 = 6

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2. limx→π(x−227)$\underset{x\to \pi }{lim}\left(x-\frac{22}{7}\right)$

Ans. limx→π(x−227)=(π−227)$\underset{x\to \pi }{lim}\left(x-\frac{22}{7}\right)=\left(\pi -\frac{22}{7}\right)$

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3. 

Ans. 

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4. 

Ans. 

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5. 

Ans. 

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6. 

Ans.  is of the form 00$\frac{0}{0}$

Put    now as 

limx→0((x+1)5−1x)$\underset{x\to 0}{lim}\left(\frac{{(x+1)}^5-1}{x}\right)$= limy→1(y5−1y−1)=limy→1(y5−15y−1)$\underset{y\to 1}{lim}\left(\frac{y^5-1}{y-1}\right)=\underset{y\to 1}{lim}\left(\frac{y^5-1^5}{y-1}\right)$             

=5⋅15−1=5⋅1=5$=5\cdot 1^{5-1}=5\cdot 1=5$                since   limx→a(xn−anx−a)=nan−1$\underset{x\to a}{lim}\left(\frac{x^n-a^n}{x-a}\right)=n{a}^{n-1}$

=5

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7. 

Ans. 

= 

= 

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8. 

Ans.   is of the form 00$\frac{0}{0}$

=

= 

= 

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9. 

Ans. 

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10. 

Ans.     is of the form 00$\frac{0}{0}$

= 

= 

= 

=  = 1 + 1 = 2

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11. 

Ans. 

= 

=  = 1

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12. 

Ans.  = 

= 

=  = 

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13. 

Ans. 

= 

= 

= ablimax→0(sinaxax)$\frac{a}{b}\underset{ax\to 0}{lim}\left(\frac{\sin ax}{ax}\right)$   [x→0⇒ax→0]$\left[x\to 0\Rightarrow ax\to 0\right]$  and [limθ→0sinθθ=1]$\left[\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1\right]$

=        

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14. 

Ans. 

=limx→0(sinaxax)ax(sinbxbx)bx=ablimx→0(sinaxax)(sinbxbx)$\underset{x\to 0}{lim}\frac{\left(\frac{\sin ax}{ax}\right)ax}{\left(\frac{\sin bx}{bx}\right)bx}=\frac{a}{b}\underset{x\to 0}{lim}\frac{\left(\frac{\sin ax}{ax}\right)}{\left(\frac{\sin bx}{bx}\right)}$            

=ab⋅limax→0(sinaxax)limbx→0(sinbxbx)$\frac{a}{b}\cdot \frac{\underset{ax\to 0}{lim}\left(\frac{\sin ax}{ax}\right)}{\underset{bx\to 0}{lim}\left(\frac{\sin bx}{bx}\right)}$ since   [x→0⇒ax→0x→0⇒bx→0]$\left[\begin{array}{c}x\to 0\Rightarrow ax\to 0\hfill \\ x\to 0\Rightarrow bx\to 0\hfill \end{array}\right]$

= ab⋅11=ab$\frac{a}{b}\cdot \frac{1}{1}=\frac{a}{b}$               [limθ→0sinθθ=1]$\left[\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1\right]$

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15. 

Ans. 

Put   now as 

 = 

=  =      [sin(−θ)=−sinθ]$\left[\sin (-\theta )=-\sin \theta \right]$

=       [limθ→0sinθθ=1]$\left[\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1\right]$

=           

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16. 

Ans.  = 

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17. 

Ans.   is of the form 00$\frac{0}{0}$

 limx→0(cos2x−1cosx−1)=limx→0(2cos2x−1)−1cosx−1$\underset{x\to 0}{lim}\left(\frac{\cos 2x-1}{\cos x-1}\right)\underset{x\to 0}{=lim}\frac{(2{\cos }^{2}x-1)-1}{\cos x-1}$       [cos2θ=2cos2θ−1]$\left[\cos 2\theta =2{\cos }^2\theta -1\right]$

 = limx→02(cos2x−1)cosx−1$\underset{x\to 0}{lim}\frac{2({\cos }^{2}x-1)}{\cos x-1}$

= 2limx→0(cosx−1)(cosx+1)(cosx−1)$2\underset{x\to 0}{lim}\frac{(\cos x-1)(\cos x+1)}{(\cos x-1)}$

= 2limx→0(cosx+1)=2(1+1)=2×2=4$2\underset{x\to 0}{lim}(\cos x+1)=2(1+1)=2\times 2=4$

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18. 

Ans. 

= 1blimx→0x(a+cosx)sinx$\frac{1}{b}\underset{x\to 0}{lim}\frac{x(a+\cos x)}{\sin x}$

= 1blimx→0xsinx⋅limx→0(a+cosx)$\frac{1}{b}\underset{x\to 0}{lim}\frac{x}{\sin x}\cdot \underset{x\to 0}{lim}(a+\cos x)$

= 1b⋅limx→0(a+cosx)limx→0(sinxx)$\frac{1}{b}\cdot \frac{\underset{x\to 0}{lim}(a+\cos x)}{\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)}$             [limθ→0sinθθ=1]$\left[\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1\right]$

= 1b×a+11=a+1b$\frac{1}{b}\times \frac{a+1}{1}=\frac{a+1}{b}$

 

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19. 

Ans.  = 

=  =  = 0

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20. 

Ans. 

Dividing numerator and denominator by 

=                           

= limx→0(sinaxax)+bxax1+(sinbxbx)bxax$\underset{x\to 0}{lim}\frac{\left(\frac{\sin ax}{ax}\right)+\frac{bx}{ax}}{1+\left(\frac{\sin bx}{bx}\right)\frac{bx}{ax}}$

 = limax→0(sinaxax)+balim1ax→01+limbx→0(sinbxbx)⋅balimbx→01$\frac{\underset{ax\to 0}{lim}\left(\frac{\sin ax}{ax}\right)+\frac{b}{a}\underset{ax\to 0}{lim1}}{1+\underset{bx\to 0}{lim}\left(\frac{\sin bx}{bx}\right)\cdot \frac{b}{a}\underset{bx\to 0}{lim}1}$                         [x→0⇒ax→0x→0⇒bx→0]$\left[\begin{array}{c}x\to 0\Rightarrow ax\to 0\hfill \\ x\to 0\Rightarrow bx\to 0\hfill \end{array}\right]$

=  1+ba1+ba=1$\frac{1+\frac{b}{a}}{1+\frac{b}{a}}=1$                                        [limθ→0sinθθ=1]$\left[\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1\right]$

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21. 

Ans. Given:

=  = 

=  = 

=  = 0