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351.

If `x_1=3 and x_ +1= sqrt(2+x_n), n ge 1`, then `lim_(nto oo) x_n` is equal toA. -1B. 2C. `sqrt(5)`D. 3

Answer» Correct Answer - B
Discussion
352.

If `f(x) =x(e^([x]+|x|)-2)/([x]+|x|)`, then `lim_(xrarr0)f(x)` is.A. -1B. 0C. 1D. non-existent

Answer» Correct Answer - D
We have,
`lim_(xto0^-)f=(x)lim_(xto0^-) x(e^(-1-x)-2)/(-1-x)[because [x]=-1and |x|=-x " for" -1 lt x lt 0]`
`rArr lim_(xto0^-)f(x)=0xx(e^-1-2)/-1=0`
and , ` lim_(xto0^+)f(x)=lim_(xto0)x((e^-x-2)/(x)) [because [x]=0 and |x|=x "for"-1lt x lt 0]`
`rArr lim_(xto0^+)f(x)=lim_(xto0)e^x-2=1-2=-1`
`therefore lim_(xto0)f(x)` does not exist.
Discussion
353.

Write the value of `(lim)_(x->-oo)(3x+sqrt(9x^2-x))`A. `1//3`B. `1//6`C. ` -1//6`D. `-1//3`

Answer» Correct Answer - B
Discussion
354.

`lim_(n->oo) {1/1.3+1/3.5+1/5.7+.....+1/((2n+1)(2n+3))` is equal toA. 1B. `(1)/(2)`C. `-(1)/(2)`D. none of these

Answer» Correct Answer - B
We have,
`lim_(ntooo) {(1)/(1.3)+(1)/(3.5)+(1)/(5.7)+......+(1)/((2n+1)(2n+3))}`
`lim_(ntooo){(1-(1)/(3))+((1)/(3)-(1)/(5))+((1)/(5)-(1)/(7))+....+((1)/(2n+1)-(1)/(2n+3))}`
` =(1)/(2) lim_(xto oo) {1-(1)/(2n+3)}=(1)/(2)lim_(xto oo) (2n+2)/(2n+3)=(1)/(2)`
Discussion
355.

If `phi(x)=lim_(x->oo)(x^(2n)(f(x)+g(x)))/(1+x^(2n))` then which of the following is correctA. `phi (x)=g(x) "for all"x in R`B. `phi (x)=f(x)"for all" x in R `C. `phix={(g(x)"for" -1ltxlt1,,),(f(x)"for"|x|ge1,,):}`D. `phix={(g(x)"for" |x|lt1,,),(f(x)"for"|x|gt1,,),((f(x)+g(x))/(2)"for"|x|=1,,):}`

Answer» Correct Answer - D
We have,
`lim_(xtooo)x^(2n)={(0if|x|lt1,,),(ooif|x|gt1,,),(1if|x|=1,,):}`
Thus, we have the following cases: ,Brgt CASE I When `-1lt x lt 1`
In this case, we have `lim_(x^2n=0`.
` therefore phi(x)=lim_(nto oo) (x^2nf(x)+g(x))/(1+x^2x)=g(x)`
CASE II When `|x|gt 1`
In this case, we have `lim_(nto oo) (1)/(x^2n)=0`
`therefore phi(x)=lim_(xtooo) (x^2nf(x)+g(x))/(1+x^2n)`
`rArr phi(x)=lim_(xtooo) (f(x)+(g(x))/(x^(2n)))/(1+(1)/(x^(2n)))=(f(x)+0)/(1+0) =f(x)`
CASE III When `|x|=1`
In this case, we have `x^(2n)=1`.
`therefore phi(x)=(f(x)+g(x))/(2)` .
Discussion
356.

`lim_(nto oo)(1+2+3+...+n)/(1+3+5+....(2n-1))=`A. `1`B. `3//2`C. `1//2`D. `2`

Answer» Correct Answer - C
We have,
`lim_(nto oo)(1+2+3+...+n)/(1+3+5+....(2n-1))=lim_(nto oo)((n(n+1))/(2))/(n^2)`
`lim_(nto oo)(n(n+1))/(2n^2) =lim_(ntooo)(n^2+n)/(2n^2)=(1)/(2)`
Discussion
357.

`Lim_(x=oo) (sqrt(x^2+1)-3sqrt(x^2+1))/(4sqrt(x^4+1)-5sqrt(x^4-1)` is equal toA. 1B. 0C. -1D. none of these

Answer» Correct Answer - B
Discussion
358.

If `f(x)=|{:(sin x,cosx,tanx),(x^3,x^2,x),(2x,1,1):}|`then `lim_(xrarr0)(f(x))/(x^2)`isA. -1B. 3C. 1D. 0

Answer» Correct Answer - C
We have,
`f(x)=|{:(sin x,cosx,tanx),(x^3,x^2,x),(2x,1,1):}|`
`rArrf(x)=x|{:(sin x,cosx,tanx),(x^2,x,1),(2x,1,1):}|" "["Taking x common form " R_2]`
`rArr(f(x))/(x^2)=|{:((sin x)/(x),cosx,tanx),(x,x,1),(2,1,1):}|`
`rArrlim_(xto0)(f(x))/(x^2)=|{:(lim_(xto0)(sin x)/(x),lim_(xto0)cosx,lim_(xto0)tanx),(lim_(xto0)x,lim_(xto0)x,1),(2,1,1):}|`
`rArrlim_(xto0)f(x)=|{:(1,1,0),(0,0,1),(2,1,1):}|=1(0-1)-(0-2)=1`.
Discussion
359.

`lim_(n rarr oo) (2+6+10+14+..."2n terms")/(2+4+6+8+..."n terms")=`______.A. 4B. 16C. 8D. 10

Answer» Correct Answer - C
Find the sum of 2n terms in numerator and sum of n terms in denominator and take the highest power of n as common in both the numerator and the denominator and then substitute `(1)/(n) = 0 "as" n rarr oo, (1)/(n) rarr 0`.
Discussion
360.

Evaluate the left- and right-hand limits of the function defined `f(x)={{:(1+x^(2)",if "0lexlt1),(2-x", if" xgt1):}"at "x=1. "Also, show that `lim_(xto1) f(x)` does not exist.

Answer» LHL of `f(x)` at x=1 is
`underset(xto1^(-))limf(x)=underset(hto0)f(1-h)`
`=underset(hto0)lim[1+(1-h)^(2)]`
`=underset(hto0)lim(2-2h+h^(2))=2`
RHL of f(x) at x=1 is
`underset(xto1^(+))limf(x)=underset(hto0)limf(1+h)`
`=underset(hto0)lim[2-(1+h)]`
`underset(hto0)lim(1-h)=1`
Clearly, `underset(xto1^(-))limf(x)neunderset(xto1^(+))f(x)`
So, `underset(xto1)limf(x)` does not exist.
Discussion
361.

Let `f` be a fuction definated on `(-1, 1)` by `f(x)=(cos ^-1 (1-{x}^2)sin^-1(1-{x}))/({x}-{x})^3 ,x ne 0 and {.}` is the fractional part function. Which of the following statements is correct ?A. `lim_(xto 0^+) f(x)` exists and equals `(pi)/(sqrt(2))`B. `lim_(xto0^-)f(x)` exists and equals `(pi)/(4)`C. `lim_(xto0^+)f(x)` exists and equals `(pi)/(4)`D. `lim_(xto 0^-) f(x)` exists and equals `(pi)/(2)`

Answer» Correct Answer - A::B
Discussion
362.

If `f(x) ={([x]^2+sin[x])/underset(0)([x])underset("for"[x]=0) "for" [x]ne 0` where `[x]` denotes the greatest integer function, then , `lim_(xto0)f(x)`, isA. 1B. 0C. -1D. non-existent

Answer» Correct Answer - D
We have ,
`lim_(xto0^-)[x]=-1and lim_(xto0^+)[x]=0`
`therefore lim_(xto0^-)f(x)=lim_(xto0^-)((-1)^2+sin(-1))/((-1))=-1+sin1`
and, `lim_(xto0^+)f(x)=lim_(xto0^+)[because f(x)=0 "for"0lex lt 1]`
So, `lim_(x to 0)f(x)` does not exist.
Discussion
363.

`lim_(xrarr0) (e^(x^(2))-cosx)/(x^2)` is equal toA. `3//2`B. ` 1//2`C. ` 2//3`D. none of these

Answer» Correct Answer - A
Discussion
364.

The value of `lim_(xrarr0) (sqrt(1-cosx^2))/(1-cosx)`, isA. `1//2`B. `2`C. `sqrt(2)`D. none of these

Answer» Correct Answer - C
Discussion
365.

Statement -1: `lim_(xrarr pi//2 ) (cot x-cosx)/(2x-pi^3)=(1)/(16)`statement 2 `lim_(xrarr0) (tanx-sinx)/(x^3)=(1)/(2)`A. Statement -1 is true, Statement-2 is true,, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for statement -1.C. Statement-1 is true, Statement-2 is False.D. Statement-1 is False, Statement-2 is true.

Answer» Correct Answer - A
Clearly,
`lim_(xto 0) (tanx-sinx)/(x^3) lim_(xto0) (sinx)/(x)xx(1-cos x)/(x^2)=1xx(1)/(2)=(1)/(2)`
So, statement -2 is true.
Now, `lim_(xto pi//2)(cot x-cos x)/(2x-pi^3)`
` =-(1)/(8)lim_(xtopi)(tan(pi//2-x)-sin(pi//2-x))/((pi//2-x)^3)`
`=(1)/(8)xx(1)/(2)=-(1)/(16)`
So, both the statements are true and statement -2 is a correct explanation for statement -1.
Discussion
366.

let `f(x)=lim_(n->oo) (x^(2n)-1)/(x^(2n)+1)`A. `f(x)={(1,|x|gt1,),(-1,|x|gt1,):}`B. `f(x)={(1,|x|lt1,),(-1,|x|gt1,):}`C. `f(x)` is not defined for any value of xD. `f(x)=1 " for " |x|=1`

Answer» Correct Answer - A
Discussion
367.

Evaluate `lim_(xto0^(-)) (x^(2)-3x+2)/(x^(3)-2x^(2)).`

Answer» `underset(xto0^(-))lim(x^(2)-3x+2)/(x^(3)-2x^(2)).`
`=underset(hto0)lim((-h)^(2)-3(-h)+2)/((-h)^(3)-2(-h)^(2))=-underset(hto0)lim(h^(2)+3h+2)/(h^(3)+2h^(2)`
When `hto0,` and Denominator`to0`
`:." "` L tends to `-oo`
Discussion
368.

`f(x)= {{:((|x-4|)/(2(x-4)), if x ne 4),(0,if x = 4):}` at `x = 4` is.

Answer» LHL of f(x) at x=4 is
`underset(xto4^(+))limf(x)=underset(hto0)limf(4-h)`
`=underset(hto0)lim(|4-h-4|)/(4-h-4)=underset(hto0)lim(|-h|)/(-h)`
`=underset(hto0)limh/-h=underset(hto0)lim-1`
`=-1`
`RHL " of " f(x) " at "x=4" is"`
`underset(xto4^(+))limf(x)=underset(hto0)limf(4+h)`
`=underset(hto0)lim(|4+h-4|)/(4+h-4)`
`=underset(hto0)lim(|h|)/(h)=underset(hto0)limh/h=underset(hto0)lim1`
`=1`
Discussion
369.

`lim_(xto0^(+)) (sum_(r=1)^(2n+1)[x^(r)]+(n+1))/(1+[x]+|x|+2x),` where `ninN` and `[.]` denotes the greatest integer function, equals

Answer» Correct Answer - C
`underset(xto0^(-))lim(sum_(r=1)^(2n+1)[x^(r)]+(n+1))/(1+[x]+|x|+2x)=underset(xto0^(-))lim(sum_(r=1)^(2n+1)[x^(r)]+(n+1))/(x)`
`=underset(xto0^(-))lim([x]+[x^(2)]+[x^(3)]+...+[x^(2n+1)]+(n+1))/(x)`
`=underset(xto0^(-))lim(ubrace((-1)+(-1)+...+(-1))_((n+1)"times")+(n+1))/(x)=underset(xto0^(-))lim(0)/(x)=0`
Discussion
370.

`lim_(xto0) [(sin(sgn(x)))/((sgn(x)))],` where `[.]` denotes the greatest integer function, is equal toA. `^(2n)p_(n)`B. `^(2n)C_(n)`C. `(2n)!`D. none of these

Answer» Correct Answer - A
`underset(xto0^(+))lim[(sin(sgnx))/(sgn(x))]=underset(xto0^(+))lim[(sin1)/(1)]=0`
`underset(xto0^(-))lim[(sin(sgnx))/(sgn(x))]`
`=underset(xto0^(-))lim[sin(-1)/(-1)]`
`=underset(xto0^(-))lim[sin1]`
Hence, the given limit is 0.
Discussion
371.

`lim_(xto0) [min(y^(2)-4y+11)(sinx)/(x)]` (where `[.]` denotes the greatest integer function) isA. -1B. 1C. 0D. 2

Answer» Correct Answer - B
`min(y^(2)-4y+11)=min[(y-2)^(2)+7]=7`
or`L=underset(xto0)lim[min(y^(2)-4y+11)(sinx)/(x)]`
`=underset(xto0)lim[(7sinx)/(x)]`
=[a value slightly lesser than 7]
`(|sinx|lt|x|," when "xto0)`
`=6`
Discussion
372.

Evaluate `lim_(ntooo) (1)/(1+n^(2))+(2)/(2+n^(2))+...+(n)/(n+n^(2)).`

Answer» `P_(n)=underset(ntooo)lim(1)/(1+n^(2))+(2)/(2+n^(2))+...+(n)/(n+n^(2))`
Now, `P_(n)ltunderset(ntooo)lim(1)/(1+n^(2))+(2)/(1+n^(2))+...+(n)/(1+n^(2))`
`=(1)/(1+n^(2))(1+2+3+...+n)`
`=(n(n+1))/(2(1+n^(2)))`
Also, `P_(n)gt(1)/(n+n^(2))+(2)/(n+n^(2))+(3)/(n+n^(2))+...+(n)/(n+n^(2))`
`=(n(+1))/(2(1+n^(2)))`
Thus `(n(n+1))/(2(n+n^(2)))ltP_(n)lt(n(n+1))/(2(n+n^(2)))`
or`" "underset(ntooo)lim(n(n+1))/(2(n+n^(2)))ltunderset(ntooo)limP_(n)lt(n(n+1))/(2(n+n^(2)))`
or` underset(ntooo)lim(1(1+((1))/(n)))/(2((1)/(n)+1))ltunderset(ntooo)limltP_(n)ltunderset(ntooo)lim(1(1+((1))/(n)))/(2((1)/(n^(2))+1))`
or `1/2ltunderset(ntooo)limP_(n)lt1/2`
or `underset(ntooo)limP_(n)=1/2`
Discussion
373.

`lim_(xrarr0^-)(sinx)/(sqrt(x))` is equal to

Answer» Correct Answer - D
Discussion
374.

The value of `lim_(xrarr0) (1+sinx-cosx+log(1-x))/(x^3)` , isA. `1//2`B. `-1//2`C. `0`D. `1`

Answer» Correct Answer - B
Discussion
375.

The value of `lim_(xrarr2)(e^(3x-6)-1)/(sin(2-x))`, isA. `3//2`B. `3`C. `-3`D. `-1`

Answer» Correct Answer - C
We have,
`lim_(xto2)(e^(3x-6)-1)/(sin(2-x))`
` lim_(xto2)(e^3(x-2)-1)/(3(x-2 ))xx (-3(2-x))/(sin(2-x))=1xx-3=-3`
Discussion
376.

If [x] denotes the greatest integer less than or equal to x, then evaluate `lim_(ntooo) (1)/(n^(3))([1^(1)x]+[2^(2)x]+[3^(2)x]+...+[n^(2)x]).`

Answer» We have, `underset(ntooo)lim(sum_(r=1)^(n)[r^(2)x])/(n^(3))=underset(ntooo)lim(underset(r=1)overset(n)sumr^(2)x-{r^(2)x})/(n^(3))`
Where `{.}` denotes the fractional part function
`=underset(ntooo)lim((x.(n(n+1)(2n+1))/(6))/n^(3)-underset(r=1)overset(n)sum{{r^(2)x}}/(n^(3)))`
`=x/6underset(ntooo)lim(1+(1)/(n))(2+(1)/(n))-underset(ntooo)limunderset(r=1)overset(n)sum{{r^(2)x}}/(n^(3))`
`=x/6-0" "( :. 0le{r^(2)x}lt1)`
`=x/6`
Discussion
377.

If a and b are positive and [x] denotes greatest integer less than or equal to x, then find `lim_(xto0^(+)) x/a[(b)/(x)].`

Answer» `L=underset(xto0^(+))limx/a[(b)/(x)].`
`=underset(xto0^(+))limx/a(b/x-{b/x})," "`where `{.}` represents the functional part function
`=underset(xto0^(+))lim(b/a-x/a{b/x})`
`=b/a-1/aunderset(xto0^(+))lim({b/x}}/(1/x)`
Since `0lt{b/x}lt1 " and "underset(xto0^(+))lim1/x=oo,` we have
`L= b/a-0=b/a`
Discussion
378.

The value of `lim_(xrarr0) (sinx)/(xqrt(x^2))`, isA. 1B. -1C. 0D. none of these

Answer» Correct Answer - D
Discussion
379.

If `f(x)={{:(x+(1)/(2)",",xlt0),(2x+(3)/(4)",",xge0):},"then"[lim_(xrarr0) f(x)]="(where[.] denotes the greatest integer function)"`A. `(1)/(2)`B. `(3)/(4)`C. does not existD. none of these

Answer» Correct Answer - C
`underset(xrarr0^(-))(lim)f(x)=underset(xrarr0^(-))(lim)(x+(1)/(2))=(1)/(2)`
`underset(xrarr0^(+))(lim)f(x)=underset(xrarr0^(+))(lim)(2x+(3)/(4))=(3)/(4)`
`therefore" "underset(xrarr0)(lim)f(x)` does not exist.
Discussion
380.

`lim_(xrarr0) x^8[(1)/(x^3)]`, where `[.]`,denotes the greatest integer function isA. a non-zero positive real numberB. a negative real numberC. 0D. non-existent

Answer» Correct Answer - C
For any `x in R`, we have
`x-1lt [x]le x`
` rArr (1)/(x^3)-1lt [(1)/(x^3)]le(1)/(x^3)"for all " 0 ne in R`
` rArr x^8((1)/(x^3-1))lt x^8[(1)/(x^3)]le x^5 " for all " x (ne 0) in R`
`rArr x^5-x^8lt x^8[(8)/(x^3)]le x^5" for all " 0 ne x in R`
But , `lim_(xto0) x^5-x^8=0and lim_(xto0) x^5=0`.
So, by Sandwich theorem, we have
`lim_(xto0) x^8[(1)/(x^3)]=0`.
Discussion
381.

`lim_(xrarr2)(sum_(r=1)^(n)x^r-sum_(r=1)^(n)2r)/(x-2)` is equal toA. nB. `(n-1)2^n`C. `(n-1)2^n+1`D. none of these

Answer» Correct Answer - C
We have ,
`S=lim_(xto2)(sum_(r=1)^(n)x^r-sum_(r=1)^(n)2r)/(x-2)`
`rArr S=lim_(xto2)(sum_(r=1)^(n)(x^r-2^r))/(x-2)`
` rArr S=sum_(r=1)^(n) lim_(xto2)((x^r-2^r)/(x-2))`
` rArr S=sum_(r=1)^(n)r.2^(r-1)`
` rArr S=1+2.2 +3.2^2+4.2^3+....+N.2^n-1` .....(i)
` rArr 2S=1.2+2.2^2+3.2^3+....+(n-1)2^(n-1)+n.2^n` .....(ii)
Subtracting (i) from (ii), we get
` S=-1-1.2-1.2^2-1.2^3.....-1.2^n-1+n.2^n`
`rArr S=-(1+2+2^2+.....+2^n-1)+n.2^n`
`rArr S=-((2^n-1)/(2-1))+n.2^n=-2^n+1+n.2^n`
Discussion
382.

If `f(x) =(sin (e^(x-2)-))/(log(x-1))` then `lim_(xrarr2)f(x)` is given byA. `-2`B. `-1`C. `0`D. `1`

Answer» Correct Answer - D
Discussion
383.

The value of `lim_(xrarr2^-) {x+(x-[x])^2}`, is

Answer» Correct Answer - D
Discussion
384.

`lim_(xrarr2) ((10-x)^(1//3)-2)/(x-2)` is equal toA. `(1)/(12)`B. `-(1)/(12)`C. `-(1)/(10)`D. `(1)/(10)`

Answer» Correct Answer - B
Discussion
385.

If `lim_(xrarroo) {ax-(x^2+1)/(x+1)}=b` , a finite number, thenA. `a=1, b=1`B. `a=0,b=1`C. `a=-1,b=1`D. `b=-1,b=-1`

Answer» Correct Answer - A
Discussion
386.

`lim_(xrarroo) (sqrt(x^2+2x-1)-x)=`A. `oo`B. `1//2`C. `4`D. `1`

Answer» Correct Answer - D
Discussion
387.

If [x] denotes the greatest integer less than or equal to x,then the value of`lim_(x->1)(1-x+[x-1]+[1-x])` is

Answer» Correct Answer - C
Discussion
388.

If `[.]` denotes the greatest integer function , then `lim_(xrarr pi//2)[(x-(pi)/(2))/(cos x)]` is equal to.A. 1B. -1C. 2D. -2

Answer» Correct Answer - D
We have,
`lim_(xtopi//2)[(x-(pi)/(2))/(cos x)]=lim_(thetato0)[(theta)/cos(pi//theta)]` where `x-pi//2=theta`
` =lim_(thetato0) [-(theta)/(sin theta)]`
` =-2 [because lim_(thetato0)(theta)/(sintheta)=1and theta gt sin theta for 0 lt theta lt (pi)/(2)]`
Discussion
389.

The value of `lim_(xrarr0) (sin(sinx)-tan(sinx))/(sin^3(sinx))`, isA. `(1)/(2)`B. `-1`C. `(1)/(2)`D. `1`

Answer» Correct Answer - A
We know that `lim_(xto0) (tanx-sin)/(sin^3x)=(1)/(2)`,
`therefore lim_(xto0) (sin(sinx)-tan(sinx))/(sin^3(sinx))`
`=-lim_(xto0) (tan(sinx)-sin(sinx))/(sin^3(sinx))=(1)/(2)`
Discussion
390.

`lim_(xrarr0) [(100 tan x sin x)/(x^2)]` is (where `[.]` represents greatest integer function).A. 99B. 100C. 0D. non-existent

Answer» Correct Answer - A
In the neighbourhood of `x=0`, we have
`0lt (tanx sin x)/(x^2)lt 1`
`rArr 0lt (100 tan x sin x)/(x^2)lt 100`
Also,
`lim_(xto0) (100tanx sin x)/(x^2)=100`
`therefore lim_(xto0) [(100 tan x sin x )/(x^2)]=99`.
Discussion
391.

`lim_(xrarr0) (3 tan3x-4 tan2x-tanx)/(4x^(2)tanx)`

Answer» Correct Answer - D
`underset(xrarr0)(lim)(3tan3x-4tan2x-tanx)/(4x^(2)tanx)`
`=underset(xrarr0)(lim)(3[tan3x-tan2x]-(tan2x+tanx))/(4x^(2)tanx)`

`=underset(xrarr0)(lim)((3sinx)/(cos2x cos3x)-(sin3x)/(cosx cos2x))/(4x^(2)tanx)`
`=underset(xrarr0)(lim)((3)/(cos2xcos3x)-(3-4sin^(2)x)/(cosx cos2x))/((4x^(2))/(cosx))`
`=underset(xrarr0)(lim)(1)/(cos2xcos3x)(3cosx-(3-4sin^(2)x)cos3x)/(4x^(2))`
`=underset(xrarr0)(lim)(3(cosx-cos3x)+4sin^(2)x)/(4x^(2))`
`=underset(xrarr0)(lim)(6sinxsin2x+4sin^(2)x)/(4x^(2))`
`=underset(xrarr0)(lim)((3)/(2)(sinx)/(x)(2sin2x)/(2)+(sin^(2)x)/(x^(2)))`
`=3+1=4`
Discussion
392.

`undersetlim_(Xrarr2^(+)) {x}(sin(x-2))/((x-2)^(2))=` (where `{.}` denotes the fractional part function)

Answer» Correct Answer - C
`underset(xrarr2^(+))(lim)(sin(x-2))/((x-2)^(2))`
`" "=underset(xrarr2^(+))(lim)({x})/((x-2))(sin(s-2))/((x-2))`
`" "=underset(xrarr2^(+))(lim)(x-[x])/(x-2)=underset(xrarr2^(+))(lim)(x-2)/(x-2)=1`
Discussion
393.

`lim_(xrarroo) sqrt((x+sinx)/(x-cos x))=`

Answer» Correct Answer - B
Discussion
394.

The value of `lim_(xrarroo)(sqrt(n^2+1)+sqrt(n))/(4sqrt(n^4+n)+4sqrt(n))`, is

Answer» Correct Answer - B
Discussion
395.

`lim_(xrarr oo) (logx)/([x])` , where `[.]` denotes the greatest integer function, is

Answer» Correct Answer - A
We have
`x-1lt[x]ge x " for all " x in R`
` rArr (1)/(x)le (1)/([x])lt (1)/(x-1) " for all " x in R -{0,1}`
` rArr (logx)/(x)le (logx)/([x]) lt (logx)/(x-1)[ because log x gt 0 " as "x to oo]`
` rArr lim_(xto oo) le lim_(xto oo) (log x)/([x])lt lim_(xle oo) (logx)/(x-1)`
` rArr lim_(xto oo) (log x)/([x]) =0 [because lim_(xto oo) (logx)/(x) =lim_(xto oo) (log x)/(x-1)=0]`
Discussion
396.

If [x] denotes the greatest integer less than or equal to x then `lim_(n->oo)([x]+[2x]+[3x]+.....+[nx])/n^2`A. `x//2`B. `x//3`C. `x`D. 0

Answer» Correct Answer - A
For ant integer k, we have
` kx-1 lt [kx]le kx`
`rArr Sigma_(k=1)^(n)(kx-1)lt Sigma_(k=1)^(n)[kx] le Sigma _(k=1)^(n)kx`
`rArr (1)/(n^2)Sigma_(k=1)^(n)(kx-1)lt (1)/(n^2)Sigma_(k=1)^(n) [kx] le (1)/(n^2) Sigma _(k=1)^(n) kx`
` rArr (x)/(n^2)Sigma_(k=1)^(n) k-(1)/(n^2)lt (1)/(n^2) Sigma _(k=1)^(n) [kx] le (x)/(n^2)Sigma _(k=1)^(n) k`
` rArr (x)/(2) (1+(1)/(n)) -(1)/(n^2) lt (1)/(n^2) Sigma _(k=1) ^(n) [kx] le (x)/(2) (1+(2)/(n))`
Now ,
` lim_(xto oo) {(1+(1)/(n))-(1)/(n^2)}=(x)/(2) and, lim_(nto oo) (x)/(2) (1+(1)/(n))=(x)/(2)`
` therefore lim_(xto oo) (1)/(n^2)Sigma _(k=1)^(n)[kx]=(x)/(2)" "["Using Sandwich Theorem "]`
i.e., `lim_(xtooo) ([x]+[2x]+[3x]+.....+[nx])/(n^2)=(x)/(2)`
Discussion
397.

`lim_(xrarr0) (x)/(tan^-1x)` is equal to

Answer» Correct Answer - B
Discussion
398.

`lim_(xrarroo) (cot^(-1)(sqrt(x+1)+sqrtx))/(sec^(-1){((2x+1)/(x-1))^(x)})=`A. 1B. `0`C. `pi//2`D. non existent

Answer» Correct Answer - A
`underset(xrarroo)(lim)(sqrt(x+1)-sqrtx)=underset(xrarroo)(lim)(1)/(sqrt(x+1)+sqrtx)=0`
`rArr" "cot^(-1)(0)=pi//2`
Also,`" "underset(xrarroo)(lim)((2x+1)/(x-1))^(x)=underset(xrarroo)(lim)((2+(1)/(x))/(1-(1)/(x)))^(x)=oo`
`rArr" "sec^(-1)(oo)=pi//2`
Thus, required limit is 1.
Discussion
399.

`lim_(x->oo) (cot^-1(sqrt(x+1)-sqrtx))/(sec^-1{((2x+1)/(x-1))^x)}` is equal toA. 1B. 0C. `(pi)/(2)`D. does not exist

Answer» Correct Answer - A
Discussion
400.

Let `f(x)=lim_(nrarroo) (tan^(-1)(tanx))/(1+(log_(x)x)^(n)),x ne(2n+1)(pi)/(2)` thenA. `AA1ltxlt(pi)/(2),f(x)` is an identity functionB. `AA(pi)/(2)ltxltpi,` the graph of f(x) is a straight line having y intercept of `-pi`C. `AA(pi)/(2)ltxlte`, the graph of f(x) is a straight line having y intercept of `-pi`D. `AAxgte, f(x)` is a constant function

Answer» Correct Answer - A::C::D
`AA1 ltx lt(pi)/(2), tan^(-1)tanx=x`
`and 0 lt log_(e)lt log_(e).(pi)/(2)lt1`
`rArr" "f(x)=x`
`AA(pi)/(2)ltx lte, tan^(-1)tanx=x-pi`
and `0lt log_(e)xlt1`
`therefore" "(log_(e)x)^(n)=0`
`rArr" "f(x)=x-pi`
and for `x gt e, log_(e)xlt1, therefore (log_(e)x)^(n)rarroo`
`rArr" "f(x)=0`
Discussion