Class 11 Maths Ncert Solutions Ex 13.1

Class 11 Maths Ncert Solutions Chapter 13 Ex 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)....(a1an)

 

Q30.

If f(x