SOME STANDARD LIMITS QUESTIONS....... TRY THESE OUT

Q10)
Putting x = 0 , y =0 we get f(0) = 0
again putting y = -x we get f(x) + f(-x) = 0 → f(x) is odd function
f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow 0}\frac{f(x+h)+f(-x)}{h}
=\lim_{h\rightarrow 0}\frac{f\left(\frac{h}{1+x^{2}+hx} \right)}{\left(\frac{h}{1+x^{2}+hx} \right)}.\frac{1}{1+x^{2}+hx} = \frac{2}{1+x^{2}}
so
f (x) =2tan^-1x+c
f(0) = c hence c = 0
thus f(x) = 2tan^-1x
so f(1/√3) = \frac{\pi }{3}
and f ' (1 ) = 1

I just hope we havent completed someone's homework.

When so many problems are dumped at one shot, my policy is to ask the OP for his attempt before answering.

QUES 7:

16) Writing series as-

cosx2ⁿ. cosx2^n-1........cosx2 (n terms)

Using the relation-
CosA.Cos2A.Cos2²A.....Cos2^n-1A = Sin(2ⁿA)2ⁿ.Sin(A)

Which makes the series bog dwn to-

Lim(n→∞) Sin2ⁿ(x/2ⁿ)2ⁿ.Sin(x/2ⁿ)

Then, use series expansion fr --- Sinx2ⁿ (n→∞).

The resultant answer comes = Sin(x)x

I think these questions are from the MATHIIT leaflet..:)

Q17)
let P = \lim_{x\rightarrow 0}\left(\frac{(1+\left\{x \right\})^{1/\left\{x \right\}}}{e} \right)^{1/\left\{x \right\}}
= \lim_{x\rightarrow 0}\left(\frac{(1+x-[x])^{1/x-[x]}}{e} \right)^{1/x-[x]}
LHL = \lim_{x\rightarrow 0}\left(\frac{(1-h-[0-h])^{1/0-h-[0-h]}}{e} \right)^{1/0-h-[0-h]}
= \lim_{x\rightarrow 0}\left(\frac{(1-h +1)^{1/\left( -h+1\right)}}{e} \right)^{1/(-h+1)} =\frac{2}{e}

and RHL = \lim_{x\rightarrow 0}\left(\frac{(1+h-[h])^{1/h-[h]}}{e} \right)^{1/h-[h]}
= \lim_{x\rightarrow 0}\left(\frac{(1+h)^{1/h}}{e} \right)^{1/h}
= \lim_{x\rightarrow 0}\left(\frac{e^{(1/h)ln(1+h)}}{e} \right)^{1/h}
= \lim_{x\rightarrow 0}e^{\left(\frac{ln(1+h)}{h}-1 \right)^{1/h}}
= \lim_{x\rightarrow 0}e^{\left(\frac{ln(1+h)}{h^{2}} \right)}
= e^{\lim_{x\rightarrow 0}{\left(\frac{\frac{ln(1+h)}{e^{h}}}{h^{2}} \right)}}
= e^{\lim_{x\rightarrow 0}}^{\frac{ln\left(1+\frac{1+h-e^{h}}{e^{h}} \right)}{(1+h-e^{h}/e^{h})}.\left(\frac{1+h-e^{h}}{e^{h}h^{2}} \right)}
= e^{\lim_{x\rightarrow 0}}^{\frac{ln\left(1+\frac{1+h-e^{h}}{e^{h}} \right)}{(1+h-e^{h}/e^{h})}.lim_{x\rightarrow 0}\left(\frac{1+h-e^{h}}{e^{h}h^{2}} \right)}
= e^RQ (say)

here Q = lim_{h\rightarrow 0}\left(\frac{1+h-e^{h}}{e^{h}h^{2}} \right)
put h = 2t
Q = lim_{h\rightarrow 0}\left(\frac{1+2t-e^{2t}}{e^{2t}(2t)^{2}} \right)
= \frac{1}{4}lim_{h\rightarrow 0}\frac{-(e^{t}-1)^{2}+2-2e^{t}+2t}{e^{2t}.t^{2}}
= Q=\frac{1}{4}\left\{-(1)^{2} +2Q\right\}\Rightarrow 2Q=-1 \Rightarrow Q=\frac{-1}{2} & R = 1
hence RHL = e^-1/2

since LHL ≠RHL
limit doesn't exist

9 > f ( x ) = ( x ) ( e )^{- [ 2x ]} , for , x > 0

= ( x ) ( e ) ⁰ = x , for , x < 0

= 0 , for , x = 0

Clearly , L . H . L = R . H . L = Functional value at x = 0

So , f ( x ) is continuos .

As f ( x ) is continuos , hence we can directly differentiate and hence test differentiabilty .

f ' ( x ) = ( 2x ) ( e ) ^{[ - 2x ]} , for x > 0

= 1 . for x < 0

= 0 , for x = 0

Clearly , f ' ( x ) is discontinuos at x = 0 .

Hence . f ( x ) is not differentiable for x = 0 .

10 > f ( x ) = a tan ^-1 x ,

Now , from the given limit , a = 2

Hence , f ( x ) = 2 tan ^{- 1} x

So , f ( 1 / √3 ) = 2 Î / 6 = Î / 3

And f ' ( 1 ) = 2 / [ 1 + 1 ] = 1

5 > Same thing as in 1 , f ( x ) = a x + b

Here , a = 1 , as f ' ( 0 ) = 1 .

So , f ( x ) = x + b , which clearly is continuos for all x .

1 > The relation given only satisfies , f ( x ) = a x + b , where a and b are any real numbers .

I can prove it if you like ,

Given , f ' ( 0 ) = - 1

So , a = - 1

And , f ( 0 ) = 1

So , b = 1

Hence , f ( 2 ) = - 1

Q11(a)
\lim_{x\rightarrow 2}\left(\frac{xf(2)-2f(x)}{x-2} \right)
using L hospital rule
\lim_{x\rightarrow 2}\left(\frac{\left\{f(2) \right\}-2f'(x)}{1} \right)

now substitute x = 2
so limit reduces to 4 - 2 x 11 = 2

Q6) Limit = \lim_{x\rightarrow 0}\frac{a(e^{x}+xe^{x})-b.\frac{1}{1+x}+c(e^{-x}-xe^{-x})}{2xsinx+x^{2}cosx} [ using L H rule ]
= a - b + c0 = \infty unless a - b + c =0
but given limit is 2 so inderterminate form 0/0 shuld continue
so a - b + c = 0 .............................(1)
then limit , [ using LH rule ]

= \lim_{x\rightarrow 0} \frac{\left\{a(1+x)e^{x}+ae^{x}+\frac{b}{(1+x)^{2}}+c(-1)e^{-x} +c(1-x)e^{-x}.(-1)\right\}}{2sinx+2xcoxsx+2xcosx-x^{2}sinx}

= \lim_{x\rightarrow 0} \frac{\left\{a(2+x)e^{x}+\frac{b}{(1+x)^{2}} -c(2-x)e^{-x}\right\}}{2sinx+4xcoxsx-x^{2}sinx}

= \frac{2a+b-2c}0{} = \infty unless 2a + b - 2c = 0
but limit = 2
so, 2a + b - 2c = 0 .............................(2)

then limit = \lim_{x\rightarrow 0} \frac{\left\{a(2+x)e^{x}+ae^{x}-\frac{2b}{(1+x)^{3}}+c(2-x)e^{-x}+ce^{-x} \right\}}{6cosx-4xsinx-2xsinx-x^{2}cosx}

= 2a + a - 2b + 2c + c6 = 2 (given )
hence 3a - 2b + 3c = 12 .............................(3)

solving equation (1) , (2) and (3) we get
a = 3 , b = 12 , c = 9 .