Solution:

Given, `f (x) = tt { ((x + 2, x <= -1),(cx^2 , x > - 1))`

`LHL = lim_( x -> -1^-) f (x) = lim_( x -> -1^-) ( x + 2)`

` = lim_( h -> 0) (-1 - h + 2) = lim_( h -> 0) ( 1 - h) = 1`

` RHL = lim_( x -> -1^+) f (x) = lim_( x -> -1^+) cx^2 = lim_( h -> 0) c ( - 1 + h)^2`

` :. = c`

If `lim_( x -> -1) f(x)` exist, then `LHL = RHL`

`:. c = 1`

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Solution:

Given, `f(x) = {((x^2 - 1, 0 < x < 2),( 2x + 3, 2 <= x < 3 ))`

`:. lim_(x -> 2^-) f(x) = lim_(x -> 2^-) (x^2- 1)`

` = lim_(h -> 0) [(2- h)^2- 1] = lim_(h -> 0) (4+ h^2 - 4h - 1)`

` = lim_(h -> 0) (h^2 - 4h + 3) = 3`

and ` = lim_(x -> 2^+) f(x) = lim_(x -> 2^+) (2x + 3)`

`= lim_(h -> 0) [2(2 + h)+ 3] = lim_(h -> 0) (4+ 2h+ 3) = 7`

So, the quadratic equation whose roots are `3` and `7` is `x^2 - (3 + 7)x + 3 xx 7 = 0 ` i.e.,

` x^2 - 10x + 21 = 0` .

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Solution:

Where `x <= 1, f(x) = x^2 - 1`

`:. lim_(x -> 1^-) f(x) = 0`

When `x > 1, f(x) = - x^2 - 1`

` :. lim_(x -> 1^+) f(x) = -2`

`:. lim_(x -> 1^-) f(x) != lim_(x -> 1^+) f(x)`

`= lim_(x -> 1) f(x)` does not exist.

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Solution:

Given, ` f (x) = tt{ (( k cos x)/(pi - 2x) , text(when) , x != pi/2 ) ,( 3, text(when) , x = pi/2) )`

` :. LHL = lim_( x -> pi^-/2) ( k cos x)/(pi - 2x) = lim_( h -> 0) ( k cos ( pi/2 - h))/( pi - 2 ( pi/2 - h))`

` = lim_( h -> 0) ( k sin h)/( pi - pi + 2h) = lim_( h -> 0) ( k sin h)/(2h)`

`= k/2 lim_( h -> 0) ( sin h)/h = k/2 . 1 = k/2`

` RHL = lim_( x -> pi^+/2) ( k cos x)/(pi - 2x) = lim_( x -> (pi/2)^+) ( k cos ( pi/2 + h))/( pi - 2 ( pi/2 + h))`

` = lim_( h -> 0) ( - k sin h)/( pi - pi - 2h) = lim_( h -> 0) ( k sin h ) /(2h) = k/2 lim_( h -> 0) ( sin h)/(2h) = = k/2 ` and ` f (pi/2) = 3`

It is given that,` lim_( x -> x//2) f (x) = f (pi/2) => k/2 = 3`

`:. k = 6`

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Solution:

`lim_(x->0) (sin (pi cos^2 x) )/x^2` (`0/0` form)

`= lim_(x->0) ( { cos (pi cos^2 x) * pi * 2 cos x (- sin x) } )/(2x)`

(using L'Hospital's rule)

`= lim_(x->0) pi cos (pi cos^2 x ) * cos x * ((- sin x)/x)`

`= (pi) * (-1) * 1 * (-1) = pi`

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Solution:

Constder `lim_(x -> 0) (sin 3x^(2))/ ( ln cos(2x^(2) - x))` `[0/0 text (from)]`

By using L'Hospital's rule, we have

`lim_(x -> 0) (cos 3x^(2) xx 6x)/ ( - sin (2x^(2) - x))/( cos (2x^(2) - x)) (4x-1)`

`lim_(x -> 0) (6x cos 3x^(2) )/(- (4 x -1) tan (2x^(2) - x) )` `[0/0 text (from)]`

Again, by using L'Hospital's rule, we have

`lim_(x -> 0) (6x cos 3x^(2) - 6x sin 3x ^(2) xx 6x )/( (-4) tan (2x^(2) - x) - (4x -1)^(2) sec^(2) (2x^(2) - x) )`

` = (6-0) /(-( -1)^(2)) = -6`

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Solution:

We have,

`lim_(x->0) (ax e^x - b log (1+x))/x^2 =3 [0/0` form `]`

Using L' Hospital's rule, we get

`lim_(x->0) (ae^x + axe^x - b/(1+x))/(2x) =3 [0/0` form `]`

`=> a-b = 0 => a= b`

Using L' Hospital's rule, we get

`lim_(x->0) (ae^x + ae^x + axe^x + b/(1+x)^2)/2 = 3`

`=> lim_(x->0) 2ae^x + axe^x + b/(1+x)^2 = 6`

`=> 2a +b = 6`..........(i)

`=> 3a = 6 => a=2`

On putting the value of a in Eq. (i), we get `b = 2`

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Solution:

Given function, ` f(x) = ( 1 - sin x)/(pi - 2x)^(2)`

where `x != pi/2 ` and ` f (pi/2) = lamda`

` lim_(x -> pi/2) f(x) = lim_(x -> pi/2) ( 1 - sin x)/(pi - 2x)^(2)`

Use 'L' Hospital rule,

` = lim_(x -> pi/2) (-cos x)/(2(pi-2x)(-2))`

` = lim_(x -> pi/2) (cos x)/(4 (pi - 2x))`

Again, use 'L' Hospital rule,

` = lim_(x -> pi/2) (- sin x) /(4 (-2)) = lim_(x -> pi/2) (sin x)/8`

` 1/8 . sin (pi/2) = 1/8 xx 1 = 1/8`

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Solution:

`lim_(theta-> 0) (4 theta(tan theta - 2 theta tan theta))/(1- cos 2 theta)`

`= lim_(theta->0) (4 (theta tan theta - 2 theta^2 tan theta))/(1- cos 2 theta)` using L' Hospital rule

`=lim _(theta-> ) (4 ( theta sec ^2 theta + tan theta - 4 theta tan theta- 2 theta^2 sec^2 theta))/(2 sin 2 theta)`

Again using L' Hospital's rule


`= lim_(theta->0) (4(sec^2 theta + 2 theta sec ^2 theta tan theta + sec ^2 theta - 4 tan theta
-4 theta sec^2 theta - 4 theta sec^2 theta- 4 theta^2 sec^2 theta tan theta))/(4 cos 2 theta)`

`=(4(1 + 0 +1))/4 =2`

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Solution:

` :. f(x) = { tt ( ( (x^3 -3x +2)/( (x-1)^2) , AA x ne 1 ) , (k, AA x =1) )`

and `f(x)` is continuous for all `x`.

`:. lim_(x->1) f(x) = f(1)`

`=> lim_(x->1) (x^3 -3x +2)/( (x-1)^2 ) = k` (`0/0` form )

By L' Hospital rule,

`k = lim_(x->1) (3x^2 -3)/(2 (x-1) )` (`0/0` form)

By L' Hospital rule,

`=lim_(x->1) (6x)/2 =3 * (1)`

`=3`

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Solution:

Given, `lim _( x -> pi) ( 1 - sin (x/2))/( cos (x/2) ( cos (x/4) - sin (x/4)))`

` = lim _( x -> pi) ( cos^2 (x/4) + sin^2 (x/4) - 2 . sin (x/4) . cos (x/4))/( cos (x/2). ( cos (x/4) - sin (x/4))) `

`quad [ ∵ sin^2 theta + cos^2 theta = 1 sin2 theta = 2sin theta cos theta]`

` = lim _( x -> pi) ( cos (x/4) - sin (x/4))^2/ (( cos^2 (x/4) - sin^2 (x/4)) ( cos (x/4) - sin (x/4))) `

`quad [ ∵ cos^2 2theta = cos^2 theta - sin^2 theta]`

` = lim _( x -> pi) ( cos (x/4) - sin (x/4))/(( cos (x/4) + sin (x/4)) ( cos (x/4) - sin (x/4))) `

`quad [ ∵ a^2 -b^2 = ( a + b) ( a -b) ]`

` = lim _( x -> pi) 1/( cos (x/4) + sin (x/4)) = 1/ ( 1/sqrt(2) + 1/sqrt(2) ) = sqrt(2) /2 = 1/sqrt(2)`

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Solution:

Given, `lim_( x -> 0) ( 2 sin x - sin 2x) /x^3 = lim_( x -> 0) ( 2 sin x - 2 sin x cos x ) /x^3 quad [ ∵ sin2x = 2 sinx cosx]`

` = lim_( x -> 0) ( 2 sin x ( 1 - cos x)) /x^3`

` = 2 lim_( x -> 0) (sin x)/x . lim_( x -> 0) ( (1 - cos x)/x^2)`

` = 2 . 1 lim_( x -> 0) (1 - cos x)/x^2 quad \ \ \ [ ∵ lim_( x -> 0) (sin x)/x = 1 ]`

` = 2 lim_( x -> 0) ( 1 - 1 + 2 sin^2 (x/2))/x^2 = 2 lim_( x -> 0) ( 2 sin^2 (x/2))/( 4 xx x^2/4)`

` = ( 2.2)/4 lim_( x -> 0) ( (sin (x/2))/(x/2))^2 = lim_( x -> 0) ( (sin (x/2))/(x/2))^2 = 1`

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Solution:

`lim_(x->oo) cos \ (sin x)/x`

Since `cos x` is continuous in `R`

`= cos [ lim_(x->oo) (sin x)/x]`

`= cos { lim_(x->oo) ("values in " [-1,1])/x}`

`= cos 0`

`=1`

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Solution:

Given, `lim_( x -> 0) ( 2 sin x - sin 2x) /x^3 = lim_( x -> 0) ( 2 sin x - 2 sin x cos x ) /x^3 quad [ ∵ sin2x = 2 sinx cosx]`

` = lim_( x -> 0) ( 2 sin x ( 1 - cos x)) /x^3`

` = 2 lim_( x -> 0) (sin x)/x . lim_( x -> 0) ( (1 - cos x)/x^2)`

` = 2 . 1 lim_( x -> 0) (1 - cos x)/x^2 quad \ \ \ [ ∵ lim_( x -> 0) (sin x)/x = 1 ]`

` = 2 lim_( x -> 0) ( 1 - 1 + 2 sin^2 (x/2))/x^2 = 2 lim_( x -> 0) ( 2 sin^2 (x/2))/( 4 xx x^2/4)`

` = ( 2.2)/4 lim_( x -> 0) ( (sin (x/2))/(x/2))^2 = lim_( x -> 0) ( (sin (x/2))/(x/2))^2 = 1`

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Solution:

Here, degree of numerator and
denominator being same.
So, we divide both by `x^3` , we get

`lim_(x->oo) (1- 3/x^2 + 2/x^3 )/( 2+ 1/x^2 - 3/x^2 ) = 1/2`

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Solution:

Now, in the limit, we get the indeterminate form `(0)(-oo)`. L’ Hospital’s Rule won’t work on products, it only works on quotients. However, we can turn this into a fraction if we rewrite things a little.

`lim_(x-> 0^(+)) x ln x = lim_(x->0^(+)) (ln x)/(1//x)`

`lim_(x-> 0^(+)) x ln x = lim_(x-> 0^(+)) (lnx)/(1//x) = lim_(x-> 0^(+)) (1//x)/(1//x^2)`

`lim_(x-> 0^(+)) x ln x = lim_(x-> 0^(+)) (1//x)/(1//x^2) = lim_(x-> 0^(+))(-x) =0`

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Solution:

This is also a problem of `oo.0`

`lim_(x-> -oo) xe^x = lim_(x->-oo) (e^x)/(1//x)`

`lim_(x->-oo) x e^x = lim_(x->-oo) (e^x)/(1//x) = lim_(x->-oo) (e^x)/(-1//x^2) = lim_(x->-oo) (e^x)/(2//x^3) = lim_(x->-oo) (e^x)/(-6//x^4)=........`

This doesn’t seem to be getting us anywhere. With each application of L’Hospital’s Rule we just end up with another 0/0 indeterminate form.

`lim_(x->-oo) xe^x= lim_(x->-oo) x/(1//e^x) =lim_(x->-oo) x/(e^(-x))`

Now by L'hospital ` =lim_(x->-oo) 1/((e^(-x)) =0`

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Solution:

Here, `lim_(x->1) (1+ sin pi x)^(cot pi x)` (`1^oo` form )

`= e^(lim_(x->1) sin pi x * cot pi x)`

`= e^(lim_(x->1)cos pi x)`

`= e^(-1)`

Hence, (b) is the correct answer.

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Solution:

`lim_(x->oo) ((x+a)/(x+b))^(x+b) = lim_(x->oo) (1+(a-b)/(x+b))^(x+b)`



`= e^(lim_(x->oo )(1+(a-b)/(x+b) - 1 ).(x+b)`


` = e^(a-b)`

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Solution:

`Lt_(x->0) ((sin x)/x)^((sin x)/(x- sin x))`

`= Lt_(x->0) [1- (1- (sin x)/x) ]^((sin x)/(x- sin x))`

Putting `y = 1 - (sin x)/x = (x- sin x)/x`

the given limit becomes

`Lt_(y->0) [(1-y)^(1/y)]^(Lt_(x->0) (sin x)/x)`

`= (e^(-1))^1 =1/e`

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Solution:

We have the indeterminate form `0^0`.

Let `y = x ^x`

taking `log` both side

`ln y = ln (x^ x) = x ln x`. Let us now find the limit of ` ln y`

`lim_(x→0^+) ln y = lim_(x→0^+) x ln x`

The above limit has the indeterminate form `0^ ∞`. We have convert it as follows

`lim_(x→0^+) x ln x`

`= lim_(x→0^+) ln x // (1 // x)`

It now has the indeterminate form `∞ // ∞` and we can use the L'hopital's theorem

`ln y = lim_(x→0^+) ln x // (1 // x)`

`= lim_(x→0^+) (1 // x) // (- 1 // x^ 2)`

`= lim_(x→0^+) -x = 0 `

The limit of `ln y = 0` and the limit of `y = x^x` is equal to

`lim_(x->0^+) y = e^ 0 = 1`

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Solution:

Let `p = lim_(x->0) (cosec x)^(1//logx)`

`=> log p=lim_(x-> 0) 1/(log x) (log cosec x)`

`= lim_(x-> 0) (log (cosec x))/(log x)`

`:. log p =lim_(x->0) ((-x cosec x cot x)/(cosec x))`

(by L' Hospital's rule)

`lim_(x-> 0) (-x/(tan x)) =-1`

`:. p=e^(-1)=1/e`

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Solution:

`lim_(x-> oo) ( sqrt (a^2 x^2 +ax +1) - sqrt (a^2 x^2 +1))`

After rationalisation,

`= lim_(x-> oo) ( (a^2x^2 +ax +1 - a^2x^2 -1)/( sqrt (a^2 x^2 +ax +1) + sqrt (a^2 x^2 +1)))`

`= lim_(x->oo) a/(sqrt (a^2 + a/x + 1/x^2) + sqrt (a^2 + 1/x^2) )`

`= (+a)/(sqrt (a^2) + sqrt (a^2) ) = a/(2a) = 1/2`

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