JEE Advanced Level Test: Limit & Derivatives- 2 - JEE MCQ

Using – L- Hospital rule

On rationalizing 

Given limit is 

By L- Hospital rule

Divide numerator and denominator by x². 

Using L hospital rule
lim x-->a d/dx[(x^⅝ - a^{^⅝})/(x^⅓ - a^⅓)
lim x-->a   [5/8(x^-⅜)]/1/3(x^-⅔)
lim x-->a 15/8 [x^⅔]/[x^⅜]
lim x-->a  15/8 [x^⅔-⅜]
Limx-->a  15/8 [x^7/24]

Using L-Hospital rule 

Using L-Hospital rule

Using L- Hospital Rule

lim(x→0) sinx sin(π/3+x) sin(π/3−x)x
= lim(x→0) (sinx/x) sin(π/3+x) sin(π/3−x)
= 1 (sin(π/3) (sin(π/3)
= 1(√3/2)(√3/2)
= 3/4

Using L-Hospital rule

Use L- Hospital Rule

Given limit 

Given limit is 

(1.2.3.....n)^1/n

By using Sandwitch theorem

 [x] can be written as X+{x}    --> {x} is fractional part .. between 0 and 1
therefore ,
lim n→∞ {[x]+[2x]+[3x]+.........+[nx]}/n²
is = lim n→∞  {x+2x+3x+.........+nx + {x}+{2x} ...+{nx}}/n²
= {x(1+2+3+..+n)   +   {x}+...{nx}}/n²
= x(n(n+1))/2n2 + ({x}+...{nx})/n²
Since {} is only b/w 0 and 1  , the second operand becomes 0 as n tends to ∞
on solving the first part , u get (n²x +nx)/2n² = x/2 + x/2n
x/2n becomes 0 as x tends to infinity.
therefore the answer is x/2.

Using L- hospital rule

Take common higher power of x in both numerator and denominator

limx→0 [(1−cos2x)(3+cosx)]/xtan4x
= limx→0(1-(1-2sin^2x))(3+cosx)/xtan4x
= limx→0 (2sin/xx)(sinx)(3+cosx)/tan4x
= lim x→0 (2sinx/x)(sinx)(3+cosx)/(4x(tan4x/4x))
= limx→0(2sinxx)(sinx)3+cosx4xtan4x4x
We know, limx→0 sinx/x=1 and limx→0 tanx/x=1
∴ our expression becomes,
= 2/4 limx→0 (3+4cosx)
= 2/4limx→0 (3+4cosx)
As cos0=1
∴ our expression becomes, =2/4(3+1) = 2

Using standard formulae

1^∞ form

e^6-1 = e⁵

Use L- Hospital rule