NCERT Solutions For Class 11 Maths Chapter 13

NCERT Solutions Class 11 Maths Limits and Derivatives

For students of class 11 who are looking to improve their knowledge on the chapter, limits and derivatives, we have provided NCERT Solutions for Class 11 Maths Chapter 13. The NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives contains topics such as limits of polynomials and rationals functions, logarithmic, trigonometric as well as exponential functions and more concepts of limits and derivatives. Thus here we have provided downloadable forms of NCERT Solutions for Class 11 Maths Chapter 13 pdf so that students can access the content offline.

NCERT Solutions Class 11 Maths Chapter 13 Exercises

Exercise 13.1

Q1.

Evaluate limx3x+5 limit:

Soln:

limx3x+5=3+5=8

 

Q2.

Evaluate limxπ(x227)

Soln:

limxπ(x227)=(π227)

 

Q3.

Evaluate limrlπr2

Soln:

limrlπr2=π(l)2=π

 

Q4.

Evaluate limx+44x+6x3

Soln:

limx+44x+6x3=4(4)+643=16+61=221=22

 

Q5.

Evaluate limx1x12+x7+1x1

Soln:

limx1x12+x7+1x1=(1)12+(1)7+111=11+12=12

 

Q6.

Evaluate limx0(x+3)53x

Soln:

limx0(x+3)53x

Put x + 3 = y so that y1asx0

Accordingly, limx0(x+3)53x=limy1y53y3=limy1y535y3=5.351=405limx0(x+3)53x=405

 

Q7.

Evaluate limx23x2x10x24

Soln:

At x = 2, the given rational function has the value of 0/ 0

limx23x2x10x24=limx2(x2)(3x+5)(x2)(x+2)=limx23x+5x+2=3(2)+52+2=114

 

Q8.

Evaluate limx3x4812x25x3

Soln:

At x = 2, the given rational function has the value of 0/ 0

limx3x4812x25x3=limx3(x3)(x+3)(x2+9)(x3)(2x+1)=limx3(x+3)(x2+9)2x+1=(3+3)(32+9)2(3)+1=6×187=1087

 

Q9.

Evaluate limx0ax+bcx+1

Soln:

limx0ax+bcx+1=a(0)+bc(0)+1=b

 

Q10.

Evaluate limz1z131z161

Soln:

limz1z131z161

At z = 1, the given function has the value of 0/ 0

Put z16=xsothatz1asx1Accordingly,limz1z131z161=limx1x21x1=limx1x212x1=2.121=2limz1z131z161=2

 

Q11.

Evaluate limx1ax2+bx+ccx2+bx+a,a+b+c0

Soln:

limx1ax2+bx+ccx2+bx+a=a(1)2+b(1)+cc(1)2+b(1)+a=a+b+ca+b+c=1

 

Q12.

limx21x+12x+2

Soln:

limx21x+12x+2

At x = -2, the given function has the value of 0/ 0

Now,limx21x+12x+2=limx2(2+x2x)x+2=limx212x=12(2)=14

 

Q13.

Evaluate limx0sinaxbx

Soln:

limx0sinaxbx

At x = 0, the given function has the value of 0/ 0

Now,limx0sinaxbx=limx0sinaxax×axbx=limx0(sinaxax)×(ab)=ablimax0(sinaxax)=ab×1=ab

 

Q14.

Evaluate limx0sinaxsinbx,a,b0

Soln:

limx0sinaxsinbx,a,b0

At x = 0, the given function has the value of 0/ 0

Now,limx0sinaxsinbx=limx0(sinaxax)×ax(sinbxbx)×bx=(ab)×limax0(sinaxax)limbx0(sinbxbx)=(ab)×11=ab

 

Q15.

Evaluate limxπsin(πx)π(πx)

Soln:

limxπsin(πx)π(πx)

It seems xπ(πx)0

limxπsin(πx)π(πx)=1πlim(zx)1sin(πx)(πx)=1π×1=1π

 

Q16.

Evaluate limx0cosxπx

Soln;

limx0cosxπx=cos0π0=1π

 

Q17.

Evaluate limx0cos2x1cosx1

Soln:

limx0cos2x1cosx1

At x = 0, the given function has the value of 0/ 0

Now,limx0cos2x1cosx1=limx012sin2x112sin2x21=limx0sin2xsin2x2=limx0(sin2xx2)×x2sin2x2(x2)2×x24=4limx0(sin2xx2)limx0sin2x2(x2)2 =4(limx0sinxx)2(limx20sinx2x2)2=41212=4

 

Q18.

Evaluate limx0ax+xcosxbsinx

Soln:

limx0ax+xcosxbsinx

At x = 0, the given function has the value of 0/ 0

Nowlimx0ax+xcosxbsinx=1blimx0x(a+cosx)sinx=1blimx0(xsinx)×limx0(a+cosx)=1b×1(limx0sinxx)×limx0(a+cosx)=1b×(a+cos0)=a+1b

 

Q19.

Evaluate limx0xsecx

Soln:

limx0xsecx=limx0xcosx=0cos0=01=0

 

Q20.

Evaluate limx0sinax+bxax+sinbxa,ba+b0

Soln:

At x =0, the given function has the value of 0/ 0

Now,limx0sinax+bxax+sinbxlimx0(sinaxax)ax+bxax+bx(sinbxbx)= =(limax0sinaxax)×limx0(ax)+limx0bxlimx0ax+limx0bx(limbx0sinbxbx)=limx0(ax)+limx0bxlimx0ax+limx0bx=limx0(ax+bx)limx0(ax+bx)=limx0(1)=1

 

Q21.

Evaluate limx0(cscxcotx)

Soln:

At x =0, the given function has the value of ∞ – ∞

Now,limx0(cscxcotx)=limx0(1sinxcosxsinx)=limx0(1cosxsinx)=limx0(1cosxsinx)(sinxx)=limx01cosxxlimx0sinxx=01=0

 

Q22.

limxπxtan2xxπ2

Soln:

limxπxtan2xxπ2

At x=π2, the given function has the value of 0/ 0

Now,putxπ2=ysothatxπ2,y0limxπ2tan2xxπ2=limy0tan2(y+π2)y=limy0tan(π+2y)y=limy0tan2yy=limy0sin2yycos2y=limy0(sin2y2y×2cos2y)=(lim2y0sin2y2y)×limy0(2cos2y)=1×2cos0=1×21=2

 

Q23.

Find limx0f(x)andlimx0f(x),wheref(x)={2x+33(x+1)x0x>0

Soln:

Given function f(x)={2x+33(x+1)x0x>0

 

limx0f(x)=limx0[2x+3]=2(0)+3=3

 

limx0f(x)=limx03(x+1)=3(0+1)=3

 

limx0f(x)=limx0f(x)=limx0f(x)=3

 

limx1f(x)=limx13(x+1)=3(1+1)=6

 

limx1f(x)=limx13(x+1)=3(1+1)=6

 

limx1f(x)=limx1f(x)=limx1f(x) = 6

 

Q24.

Find limx1f(x),wheref(x)={x21,x21,x1x1

Soln:

Given function is

f(x)={x21,x21,x1x1 limx1f(x)=limx1[x21]=121=11=0 limx1+f(x)=limx1[x21]=121=11=0

We observed that limx1f(x)limx1+f(x)

Hence, limx1f(x) doesn’t exist.

 

25.

Evaluate limx0f(x),wheref(x)={|x|x,0,x0x=0

Soln:

Given function f(x)={|x|x,0,x0x=0

When |x|=x

limx0f(x)=limx0[|x|x]

= limx0(xx)

= limx0(1)

= -1

When |x|=x

limx0+f(x)=limx0+[|x|x]

= limx0[xx]

= limx0(1)

= 1

We observe that limx0f(x)limx0+f(x)

Hence, limx0f(x) doesn’t exist.

 

Q26.

Find limx0f(x),wheref(x)={x|x|,0,x0x=0

Soln:

f(x) = {x|x|,0,x0x=0

When |x|=x

limx0f(x)=limx0[x|x|] =limx0[xx] =limx0(1)

= -1

When |x|=x

limx0+f(x)=limx0+[x|x|] =limx0[xx] =limx0(1)

= 1

We observe that limx0f(x)limx0+f(x)

Hence, limx0f(x) doesn’t exist.

 

Q27.

Find limx5f(x),wheref(x)=|x|5

Soln:

Given function is f(x)=|x|5

When x>0,|x|=x

limx5f(x)=limx5[|x|5]

= limx5(x5)

= 5 – 5

= 0

When x>0,|x|=x

limx5+f(x)=limx5+(|x|5)

= limx5(x5)

= 5 – 5

= 0

limx5f(x)=limx5+f(x)=0

Hence, limx5f(x)=0

 

Q28.

Suppose f(x)=a+bx,4,bax,ifx<1ifx=0,ifx>1

And limx1f(x)=f(1) what can be the values of b and a?

Soln:

Given function

f(x)=a+bx,4,bax,ifx<1ifx=0,ifx>1 limx1f(x)=limx1(a+bx)=a+b limx1f(x)=limx1(bax)=ba

f(1) = 4

Given that limx1f(x)=f(1).

limx1f(x)=limx1+f(x)=limx1f(x)=f(1)

=> a+ b = 4 and b – a = 4

Solving both the equations, we get a = 0 and b = 4.

Hence, b and a are 4 and 0

 

Q29.

A function f(x)=(xa1)(xa2)....(xan) is define by fixed real numbers a1, a2, . . . . an.

Find limxa1f(x). For some aa1,a2,....,ancomputelimxanf(x)

Soln:

Given function = f(x)=(xa1)(xa2)...(xan)

limxa1f(x)=limxa1[(xa1)(xa2)...(xan)]

 

=[limxa1(xa1)][limxa1(xa2)]...[limxa1(xan)]

 

=(a1a1)(a1a2)....(a1an)=0

 

limxa1]f(x)=0

 

Now, limxa1]f(x)=limxa[(xa1)(xa2)...(xan)]

 

= =limxa[xa1][xa2]...[xan]

 

(a1a1)(a1a2)....(a1an)
limxaf(x)=(a1a1)(a1a2)....(a1