Limits and Continuity Exercise 16.2 | Basic Mathematics Solution [NEB UPDATED]

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Exercise 16.2

Evaluate the following:

1) x lim0 sinaxx

Solution:

x lim0 sinaxx

When x = 0, the given function takes the form 00.

=x lim0 sinaxx = x lim0 sinaxax.a = 1.a = a.

 

2) x lim0 tanbxx

Solution:

x lim0 tanbxx

When x = 0, the given function takes the form 00.

=x lim0 tanbxx = x lim0 sinbxx.cosbx.a = x lim0 sinbxbx.b.1cosbx = 1.b.1 = b.

 

3) x lim0 sinmxsinnx

Solution:

x lim0 sinmxsinnx

When x = 0, the given function takes the form 00.

=x lim0 sinmxsinnx = x lim0 sinmxmx.m.1sinnxnx.n = 1.m.11.n = mn.

 

4) x lim0 tanaxtanbx

Solution:

x lim0 tanaxtanbx

When x = 0, the given function takes the form 00.

=x lim0 tanaxtanbx = x lim0 sinaxcosaxsinbxcosbx= x lim0 sinaxcosax.cosbxsinbx

x lim0 = sinaxax.a.1cosax.cosbxsinbxbx.b = 1.a.1.11.b = ab.

 

5) x lim0 sinpxtanqx

Solution:

x lim0 sinpxtanqx

When x = 0, the given function takes the form 00.

=x lim0 sinpxtanqx = x lim0 sinpx.cosqxsinqx= x lim0 sinpxpx.p.cosqxsinqxqx.q = 1.p.11.q = pq.

 

6) x lima sin(xa)x2a2

Solution:

x lima sin(xa)x2a2

When x = a, the given function takes the form 00.

Or, x lima sin(xa)x2a2 = x lima sin(xa)(x+a)(xa)= x lima sin(xa)xa.1x+a = {(xa)lim0sin(xa)xa}{xlima1x+a}

 = 1. 1a.a = 12a.

 

7) x limp x2p2tan(xp)

Solution:

x limp x2p2tan(xp)

When x = p, the given function takes the form 00.

=x limp x2p2tan(xp) = x limp x2p2sin(xp)cos(xp)= x limp (x+p)(xp).cos(xp)sin(xp)

=x limp xpsin(xp).(x+p).cos(xp)

= (x – p) lim0 1(sin(xp))xp. x limp (x+p).cos(xp)= 1.(p + p).1 = 2p.

 

8) x lim0 sinax.cosbxsincx

Solution:

x lim0 sinax.cosbxsincx

When x = 0, the given function takes the form 00.

=x lim0 sinax.cosbxsincx = x lim0 sinaxax.a.acosbxsincxcx.c = 1.a.11.c = ac.

 

9) x lim0 1cosbxx2

Solution:

x lim0 1cosbxx2

When x = 0, the given function takes the form 00.

=x lim0 1cosbxx2 = x lim0 2sin2x2x2 = x lim0[2.(sinx2x2)2.14].

= 12(1)2 = 12.

 

10) x lim0 1cos6xx2

Solution:

x lim0 1cos6xx2

When x = 0 , the given function takes the form 00.

=x lim0 1cos6xx2 = x lim0 2sin23xx2 = x lim0[2.(sin3x3x)2.9] = 18.

 

11) x lim0 1cos9xx2

Solution:

x lim0 1cos9xx2

When x = 0, the given function takes the form 00.

= x lim0 1cos9xx2 = x lim0 2sin29x2x2 = x lim0 2(sin9x2)2x2 = x lim0 [2(sin9x29x2)2(814)] = 812.

 

12) x lim0 cosaxcosbxx2[00form]

Solution:

x lim0 cosaxcosbxx2[00form]

= 2sinax+bx2.sinbxax2x2 = 2sin(a+b)x2x.sin(ba)x2x

= 2sin(a+b)x2(a+b2)x.2a+b.sin(ba)x2(ba2)x.2ba = 2.a+b2.ba2 = b2a22.

 

13) x lim0 sinaxsinbxx[00form]

Solution:

x lim0 sinaxsinbxx[00form]

= x lim0 (sinaxxsinbxx) = x lim0 (sinaxax.asinbxbx.b)

= 1.a – 1.b = a – b.

 

14) x lim0 1cospx1cosqx[00form]

Solution:

x lim0 1cospx1cosqx[00form]

= x lim0 2sin2px2sin2qx2 = x lim0 (sinpx2)2(sinqx2)2

= x lim0 [(sinpx2px2)2.p24.1(sinqx2qx2)q24] = p2q2.

 

15) x lim0 tanxsinxx3[00form]

Solution:

x lim0 tanxsinxx3[00form]

= x lim0 sinxcosxsinxx3 = x lim0 sin2xsin2x.cos2xx3.cos2x

= x lim0 sin2x(1cos2x)x3cos2x = x lim0 sin2x.2sin2xx3cos2x.

= x lim0 sin2x2x.2.2(sinxx)2.1cos2x = 1.4.1 = 4.

 

17) x limπ/2  (secx – tanx)  

Solution:

x limπ/2  (secx – tanx)  

= x limπ/2  (secx – tanx)   [∞ – ∞ form]

= x limπ/2 (1cosxsinxcosx) = x limπ/2 (1sinxcosx)

= x limπ/2 1sinxcosx * 1+sinx1+sinx = x limπ/2  1sin2x2cosx(1+sinx)

= x limπ/2 cosx1+sinx = cosπ21+sinπ2 = 0.

 

18) x limπ/4 sex2x2tanx1[00form]

Solution:
x limπ/4 sex2x2tanx1[00form]

=  x limπ/4 sex2x2tanx1[00form]

= x limπ/4 1+tan2x2tanx1 = x limπ/4 tan2x1tanx1.

= x limπ/4 (tanx+1)(tanx1)tanx1 = x limπ/4 (tanx + 1)

= tan π4 + 1 = 1 + 1 = 2.

 

19) x limπ/4 = 2cosec2x1cotx[00form]

Solution:

x limπ/4 = 2cosec2x1cotx[00form]

= x limπ/4 = 2cosec2x1cotx[00form]

= x limπ/4 21cot2x1cotx = x limπ/4 1cot2x1cotx

= x limπ/4 (1+cotx)(1cotx)1cotx = x limπ/4 (1 + cotx)

= 1 + cotπ4 = 1 + 1 = 2.

 

20) x limy tanxtanyxy[00form]

Solution:

x limy tanxtanyxy[00form]

= x limy tanxtanyxy[00form]

= x limy (sinxcosxsinycosyxy) = x limy sinx.cosycosx.sinycosx.cosy(xy).

= x limy sin(xy)cosx.cosy(xy).

= x limy sin(xy)xy. x limy 1cosx.cosy

= 1.1cosy.cosy = sec2y.

 

21) x limy sinxsinyxy[00form]

Solution:

= x limy sinxsinyxy[00form].

= x limy 2cosx+y2.sinxy2xy

= 2 x limy cos x+y2. x – y lim0 sinxy2xy2.2 = 2.cos2y2.12 = cosy.

 

22) x limy cosxcosyxy[00form]

Solution:
x limy cosxcosyxy[00form]

= x limy cosxcosyxy[00form]

= x limy 2sinx+y2.sinyx2xy = x limy 2sinx+y2.sinxy2xy

= –2 x limy sin x+y2. (x – y) lim0 sinxy2xy2.2

= –2sin 2y2.12 = –siny.

 

23) x limθ xcotθθcotxxθ[00form]

Solution:
x limθ xcotθθcotxxθ[00form]

= x limθ xcotθθcotxxθ[00form].

= x limθ xcotθθcotθ+θcotθθcotxxθ

= x limθ (xθ)cotθ+θ(cotθcotx)xθ

= x limθ [(cotθ+θxθ{cosθsinθcosxsinx}]

= x limθ [cotθ+θxθ.sinxcosθcosx.sinθsinθ.sinx]

= cotθ + x limθ θ(xθ).sin(xθ)sinθ.sinx

= cotθ + x – θ lim0 sin(xθ)xθ. x limθ θsinθ.sinx.

= cotθ + 1.θsinθ.sinθ = cotθ + θsin2θ.

 

24) x limθ xcosθθcosxxθ[00form]

Solution:

x limθ xcosθθcosxxθ[00form]

= x limθ xcosθθcosxxθ[00form]

= x limθ xcosθθcosθ+θcosθθcosxxθ

= x limθ (xθ)cosθ+θ(cosθcosx)xθ

= x limθ [cosθ+θxθ.2sinθ+x2.sinxθ2]

= cosθ + x limθ 2θsin θ+x2x – θ lim0 sinxθ2xθ2.2

= cosθ + 2θ.sin 2θ2.1.12 = cosθ + θ.sinθ.

 

25) x lim1 1+cosπxtan2πx

Solution:

x lim1 1+cosπxtan2πx

= x lim1 1+cosπxtan2πx = x lim1 (1+cosπx)cos2πx(1+cosπx)(1cosπx)

= cos2π1cosπ

= (1)21(1) = 12.

 

26) x limθ xtanθθtanxxθ[00form]

Solution:

= x limθ xtanθθtanxxθ[00form].

= x limθ xtanθθtanθ+θtanθθtanxxθ.

= x limθ (xθ)tanθ+θ(tanθtanx)xθ

= x limθ [tanθ+θxθ{tanθtanx}]

= x limθ [tanθ+θxθ{sinθcosθsinxcosx}]

= x limθ [tanθ+θxθ{sinθ.cosxcosθ.sinxcosθ.cosx}]

= x limθ [tanθ+θxθ.sin(θx)cosθ.cosx]

= tanθ – x limθ θcosθ.cosxx – θ lim0 sin(xθ)xθ

= tanθ –θcosθ.cosθ.1 = tanθ – θsec2θ.

 

27) θ limπ/4 cosθsinθθπ4[00form]

Solution:

= θ limπ/4 cosθsinθθπ4[00form]

= θ limπ/4 2(12cosθ12sinθ)θπ4

= θ limπ/4 2(sinπ4cosθcosπ4.sinθ)θπ4

= θ limπ/4 2sin(π4θ)θπ4

= θ π4lim0 2(θπ4)θπ4

= 2.1 = 2.

 

28) x limc xcsinxsinc[00form]

Solution:

x limc xcsinxsinc[00form]

= x limc xcsinxsinc[00form]

= x limc xc2sinxc2.cosx+c2 * x+cx+c

= x limc xc2sinxc2.cosx+c2(x+c)

= x limc 22(x+c).sinxc2xc2.cosx+c2

= x limc 1(x+x)cosx+c2 = 1(c+c).cosc+c2 = 12c.cosc.

= secc. 12c = secc2c.

 

29) Find the limit of:

a) x lim0 e6x1xMissing or unrecognized delimiter for \right

Solution:

x lim0 e6x1xMissing or unrecognized delimiter for \right

= x lim0 e6x1x(00) = (xlim0e6x1x).6 = 1.6 = 6.

 

b) x lim0 e2x1x.2x+1,(00)

x lim0 e2x1x.2x+1,(00)

= x lim0 e2x1x.2x+1,(00) = (xlim0e2x12x.22x+1) = 1.22 = 1.

 

c) x lim0 eaxebxx

Solution:

x lim0 eaxebxx

= x lim0 eaxebxx ,(00) = x lim0 eax1(ebx1)x  = x lim0 eax1x – x lim0 ebx1x = (xlim0eax1ax).a(xlim0ebx1bx).b = 1.a – 1.b = a – b.

 

d) limx0ax+bx2x

Solution:

limx0ax+bx2x

= x lim0 ax+bx2x,(00) = = x lim0 ax1x+=xlim0bx1x= loga + logb = log(ab).

 

30) Evaluate the limit: 

a) x lim2 x2log(x1)

Solutio:

x lim2 x2log(x1)

let x – 2 = y → x = y + 2.

So, x → 2 y → 0.

Now, = x lim0 x2log(x1) = ylim0 ylog(y+1) = 1ylim0log(y+1)y = 11 = 1.

 

b) $ \mathop {\lim }\limits_{x \to \frac{\pi }{2}} {\rm{ }}\frac{{\cos x}}{{\log (x - \frac{\pi }{2} + 1)}}$

Solution:

limxπ2cosxlog(xπ2+1)

= x limπ/2 cosxlog(xπ2+1), (00)

Let x – π/2 = y → y x = π2 + y

So, x →π2 y → 0

Now,

=x limπ/2 cosxlog(xπ2+1)

= y lim0 cos(π2+y)log(y+1) = y lim0 sinylog(y+1).

= –y lim0 (sinyylog(y+1)y) = 11 = –1.

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