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151.

Let `f(x) =(kcosx)/(pi-2x)` if `x!=pi/2` and `f(x=pi/2)` if `x=pi/2`then find the value of `k` if `lim_(x->pi/2) f(x)=f(pi/2)`

Answer» Given, `{{:((kcosx)/(pi-2x)"," , "When " xne (pi)/2),(3",", "When " x=(pi)/2):}`
`therefore` LHL =`lim_(xto(pi^(-))//2)(kcosx)/(pi-2x)= lim_(hto0)(kcos(pi/2-h))/(pi-2(pi/2-h))`
`k/2lim_(hto0)(sinh)/(h)=k/2.1=k/2` `[therefore lim_(hto0)(sinx)/(x)=1]`
RHL `=lim_(hto0)(kcosx)/(pi-2x)=underset(xto(pi//2))"lim"+(kcos(pi/2+h))/(pi-2(pi/2+h))`
`=lim_(hto0)(-ksinh)/(pi-pi-2h)=lim_(hto0)(ksinh)/(h)=k/2lim_(hto0)(sinh)/(2h)=k/2 "and" f(pi/2)=3`
It is given that, `therefore lim_(xto(pi//2))f(x)=f(pi/2)rArr k/2=3` k=6
Discussion
152.

Evaluate `lim_(xto(pi/3)) sqrt(1-cos6x)/(sqrt(2)(pi/3-x)`

Answer» Given, `underset(xto(pi/3))"lim"sqrt(1-cos6x)/(sqrt(2)(pi/3-x)) = underset(xto3)"lim"sqrt(1-1+2sin^(2)+3x)/(sqrt(2)(pi/3-x))` `[therefore cos2x=1-2sin^(2)x]`
`=underset(xto(pi)//3)"lim"(sqrt(2)sin3x)/(sqrt(2)(pi/3-x)=underset(xto(pi)/3)"lim"(sin3x)/(pi/3-x)` `[therefore cos2x=1-2sin^(2)x]`
`=underset(xto(pi)//3)"lim"(sqrt(2)sin3x)/(sqrt(2)(pi/3)-x) = underset(xto(pi)/3)"lim"(sin3x)/(pi/3-x)`
`=underset(xto(pi//3))"lim" (sin(pi-3x))/(pi-3x)=3 xx 1` `[therefore underset(xto0)"lim"(sinx)/(x) =1]`
=3 `[therefore xto(pi/3) rArr (x-pi/3)to 0]`
Discussion
153.

`lim_(xto(pi)) (1-sin(x/2))/(cosx/2(cosx/4-sinx/4))`

Answer» Given, `underset(xto(pi))"lim"(1-sin(x/2))/(cosx/2(cosx/4-sinx/4))`
`=underset(xto(pi))"lim"(cos^(2)x/4 + sin^(2)x/4-2.sinx/4.cosx/4)/(cosx/2.(cosx/4-sinx/4))` `[therefore sin^(2)theta+cos^(2)theta=1sin2theta=2sinthetacostheta]`
`=underset(xto(pi))"lim"(cosx/4-sinx/4)^(2)/((cos^(2)x/4-sin^(2)x/4)(cosx/4)-sin(x/4))` `[therefore cos^(2)2theta=cos^(2)theta-sin^(2)theta]`
`=underset(xto0)"lim"((cos(x/4)-sin(x/4))/((cosx/4+sinx/4)(cosx/4-sinx/4)))` `[therefore a^(2)-b^(2)=(a+b)(a-b)]`
=`underset(xto(pi))"lim"1/(cosx/4+sin x/4)=1/(1/sqrt(2)+1/sqrt(2))=sqrt(2)/2=1/sqrt(2)`
Discussion
154.

`lim_(xto(pi//4)) (tan^(3)x-tanx)/(cos(x+pi/4))`

Answer» Given, `underset(xto(pi//4))"lim"(tan^(3)x-tanx)/(cos(x+pi/4))`
`=underset(xto(pi)/4)"lim"(tanx(tan^(2)x-1))/(cos(x+pi/4))= underset(xto(pi//4))"lim"tanx.underset(xto(pi//4))"lim"((1-tan^(2)x)/(cos(x+(gpi)/4)))`
`=underset(xto(pi//4))(-1xx"lim")((1+tanx)(1-tanx))/(cos(x+pi/4))` `[therefore a^(2)-b^(2)=(a+b)(a-b)]`
`-underset(xto(pi//4))"lim"(1+tanx) underset(xto(pi//4))"lim"[(cosx-sinx)/(cosx.cos(x+pi/4)]]`
`=-(1+1) xx underset(xto(pi//4))"lim"(sqrt(2)[1/sqrt(2).cosx-1/sqrt(2).sinx])/(cosx.cos(x+pi/4))=-2sqrt(2)underset(xto(pi//4))"lim"[(cospi/4.cosx-sinpi/4.sinx)/(cosx.cos(x+pi/4))]` `[therefore cosA.cosB-sinAsinB=cos(A+B)]`
`=-2sqrt(2)underset(xto(pi//4))"lim"(cos(x+pi/4))/(cos.cos(x+pi/4))=-2sqrt(2)xx1/(1/sqrt(2))=-2sqrt(2)xx sqrt(2)=-4`
Discussion
155.

Evaluate, `lim_(xto(pi//6)) (cot^(2)x-3)/("cosec"x-2)`A. `2`B. `3`C. `4`D. `5`

Answer» Correct Answer - C
Given, `underset(xto(pi//6))"lim"(cot^(2)x-3)/("cosec"x-2) = underset(xto(pi//6))"lim"("cosec"^(2)x-1-3)/("cosec"x-2)` `[therefore"cosec"^(2)x=1+cot^(2)x]`
`=underset(xto(pi//6))"lim"("cosec"^(2)x-4)/("cosec"x-2)=underset(xto(pi//6))"lim"(("cosec"x)^(2)-(2)^(2))/("cosec"x-2)`
`=underset(xto(pi//6))"lim"(("cosec"x+2)("cosec"x-2))/("cosec"x-2)`
`="cosec"pi/6+2=2+2=4`
Discussion
156.

Evaluate the following limit: `lim_(x->oo) (sin(a+x) + sin(a-x) - 2 sin a)/(x sinx))`

Answer» `lim_(x->0) (sin(a+x)+sin(a-x)-2sina)/(xsinx)`
`=lim_(x->0) (sinacosx+sinxcosa+sinacosx-sinxcosa-2sina)/(xsinx)`
`=lim_(x->0) (2sina(cosx-1))/(xsinx)`
`=2sina lim_(x->0) (-2sin^2(x/2))/(x*2sin(x/2)cos(x/2))`
`=-2sina lim_(x->0)(sin(x/2))/(2*x/2*cos(x/2))`
As, `lim_(x->0) sinx/x = 1 and lim_(x->0) cos(x/2) = cos 0^@ = 1``=2sina*(1/(2*1)) = -sina`
Discussion
157.

`lim(xrarrpi)(x-(22)/(7))`

Answer» `underset(xrarrpi)"lim"(x-(22)/(7))=(pi-(22)/(7))`
Discussion
158.

Let `a_(1),a_(2),.......,a_(n)` be fixed real numbers and define a function `f(x)=(x-a_(1))(x-a_(2)) ......(x-a_(n))`, what is lim `f(x)`? For some `anea_(1),a_(2),.........a_(n)`, compute `lim_(Xrarr1) ` f(x)

Answer» `underset(xrarra_(1))"lim"f(x)= underset(xrarra_(1))"lim"[(x-a_(1)]`
`(X-a_(2))........(x-a_(n))`
`=(a_(1)-a_(1))(a_(1)-a_(2))......(a_(1)-a_(n ))=0`
If `anea_(1),a_(2),a_(3).........a_(n)` then
`underset(Xrarra)"lim"f(x)=underset(Xrarra)"lim"[x-a_(1)(x-a_(2))........(x-a_(n))]`
`=(a-a_(1))(a-a_(2))......(a-a_(n))`
Discussion
159.

`(lim)_(x->0)(s i n a x)/(sinb x)a ,b ,!=0`

Answer» Here, we will use the property,
`lim_(x->0)sinx/x = 1`
`lim_(x->0) (sinax)/(sinbx) = lim_(x->0)((axsin ax)/(ax))/((bxsinbx)/(bx))`
`= lim_(x->0)(ax(sin ax)/(ax))/(bx(sin bx)/(bx))`
`= lim_(x->0) a/b[((sin ax)/(ax))/((sin bx)/(bx))]`
`= a/b[(lim_(x->0)((sin ax)/(ax)))/(lim_(x->0)((sin bx)/(bx)))]`
Now, let `ax = y and bx = z`
Then, our expression becomes,
`= a/b[(lim_(y->0)((sin y)/(y)))/(lim_(z->0)((sin z)/(z)))]`
`=a/b[1/1] = a/b`
Discussion
160.

if `f(x) = {{:(x+2",",xle1),(cx^(2)",",xgt-1):},` then find c when `lim(xto-1)f(x)` exists.

Answer» Given, `f(x) = {{:(x+2",",xle1),(cx^(2)",",xgt-1):},`
LHL=`lim_(xto-1^(-))f(x)= underset(xto-1^(-1))"lim"(x+2)`
`=lim_(hto0)(-1-h+2)=lim_(hto0)(1-h)=1`
RHL `=lim_(xto-1^(+))f(x)= lim_(xto-1^(+)) cx^(2)=lim_(hto0)c(-1+h)^(2)`
`therefore` = c
If `lim_(xto-1)f(x)` exist, then LHL = RHL
`therefore c=1`
Discussion
161.

Evaluate(i) `(lim)_(x->1)(x^(15)-1)/(x^(10)-1)` (ii) `(lim)_(x->0)(sqrt(1+x)-1)/x`

Answer» Here, we will use,
`lim_(x->a) (x^n-a^n)/(x-a) = na^(n-1)`

(i)`lim_(x->1)(x^15-1)/(x^10-1)``= lim_(x->1)((x^15-1^15)/(x-1))/((x^10-1^10)/(x-1))`
`=(15(1)^14)/(10(1)^9) = 15/10 = 3/2`

(ii)`lim_(x->0) (sqrt(1+x) -1)/x`
`= lim_(1+x->1)((1+x)^(1/2) - (1)^(1/2))/((1+x)-1)`
`=1/2(1)^(-1/2) = 1/2`
Discussion
162.

Let `p=(lim)_(xvec0+)(1+tan^2sqrt(x))^(1//2x)`then `logp`is equal to:(1) 2(2) 1(3) `1/2`(4)`1/4`

Answer» `p = lim_(x->0^+) ( 1+ tan^2 sqrtx)^(1/(2x))`
`log p = lim_(x->0^+) log(1+ tan^2 sqrtx)^(1/(2x))`
`= lim_(x->0^+) 1/(2x) log(1+ tan^2 sqrtx)`
`= lim_(x->0^+) (Log(1+tan^2 sqrt x))/(2x)`
using L hospital rule
`lim_(x->0) (ax^2)/(bx^3) = (2ax)/(3bx^2)`
`= lim_(x->0^+) (1/(1+ tan^2 sqrtx) xx 2 tan sqrt xx sec^2 sqrt x xx 1/(2 sqrt x))/2`
`log p = lim_(x->0+) ( tan sqrt x xx sec^2 sqrt x)/ ( 2 sqrt x xx ( 1 + tan^2 sqrtx))`
`= (1 xx 1)/(2 xx (1+0))`
`log p = 1/2 `
option 2 is correctAnswer
Discussion