Courses

# Test: Infinite Limits

## 10 Questions MCQ Test Mathematics For JEE | Test: Infinite Limits

Description
This mock test of Test: Infinite Limits for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Test: Infinite Limits (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Infinite Limits quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Infinite Limits exercise for a better result in the exam. You can find other Test: Infinite Limits extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1
Solution:

lim (x → 0) [((1-3x)+5x)/(1-3x)]1/x
lim (x → 0) [1 + 5x/(1-3x)]1/x
= elim(x → 0) (1 + 5x/(1-3x) - 1) * (1/x)
= elim(x → 0) (5x/(1-3x)) * (1/x)
= elim(x → 0) (5x/(1-3x))
= e5

QUESTION: 2
Solution:

QUESTION: 3
Solution:

lim(x→0) [log10 + log1/10]/x
= [log10 + log10]/0
= 0/0 form
lim(x→0) [(1/(x+1/10) * 1]/1
lim(x→0) [(1/(0+1/10) * 1]/1
= 1/(1/10) => 10

QUESTION: 4

Solution:

QUESTION: 5

Solution:

lim(x → 1) (log2 2x)1/log2x
= lim(x →1) (log22 + log2x)1/log2x
As we know that {log ab = log a + log b}
lim(x → 1) {1 + log2x}1/log2x
log2x → 0
Put t = log2x
lim(t → 0) {1 + t}1/t
= e

QUESTION: 6

Solution:
QUESTION: 7

lim(x → 0) (tanx/x)(1/x^2)

Solution:

lim(x → 0) (tanx/x)(1/x^2)
= (1)∞
elim(x → 0) (1/x2)(tanx/x - 1)
= elim(x → 0) ((tanx - x)/x3)   .....(1)
lim(x → 0) ((tanx - x)/x3)
(0/0) form, Apply L hospital rule
lim(x → 0) [sec2x -1]/3x2
lim(x → 0) [tan2x/3x2]
= 1/3 lim(x → 0) [tan2x/x2]
= 1/3 * 1
= e1/3

QUESTION: 8

Solution:
QUESTION: 9

Solution:

lim(x → ∞) [(x-2)/(x+3)]2x
lim(x → ∞) [(x-2)/(x+3)]2 * [(x-2)/(x+3)]x

lim(x → a) [f(x) * g(x)] = lim(x → a) f(x) * lim(x → a) g(x)

lim(x → ∞) [(x-2)/(x+3)]2 * lim(x → ∞) [(x-2)/(x+3)]x
Lets evaluate the limits of both the functions separately,
lim(x → ∞) [(x-2)/(x+3)]2
= lim(x → ∞) [(x(1-2/x)/(1+3/x)]2
lim(x → ∞) [(1-2/x)/(1+3/x)]2
Apply infinity property,
= [(1-0)/(1-0)]2
= 1
Now, lim(x → ∞) [(x-2)/(x+3)]2
= lim(x → ∞) exln[(x-2)/(x+3)]
= lim(x → ∞) ln[(x-2)/(x+3)]/(1/x)
Apply L hospital rule
lim(x → ∞) d/dx[ln(x-2)/(x+3)]/[d/dx(1/x)]
= lim(x → ∞) {1/[(x-2)/(x+3)] * d/dx[(x-2)/(x+3)]}/(-1/x2)
= lim(x → ∞) {(x+3)/(x-2)[(x+3)d/dx(x-2) - (x-2)d/dx(x+3)]/(x+3)2}/(-1/x2)
= lim(x → ∞) {(x+3)/(x-2)[(x+3)-(x-2)]/(x+3)2}/(-1/x2)
= lim(x → ∞) {(x+3)/(x-2)[5/(x+3)2]}/(-1/x2)
= lim(x → ∞) [-5x2/(x+3)(x-2)]
= lim(x → ∞) [-5x2/(x2 + x - 6)]
Again apply L hospital rule,
= lim(x → ∞) [d/dx(-5x2)/(d/dx(x2 + x - 6))]
= lim(x → ∞) [-10x/(2x + 1)]
Again applying L hospital rule,
= lim(x → ∞) [d/dx(-10x)/(d/dx(2x + 1))]
=  lim(x → ∞) [-10/2]
=  lim(x → ∞) [-5]
= -5
= lim(x → ∞) exln[(x-2)/(x+3)]2 = e(-5)2
= 1/e10
= lim(x → ∞) [(x-2)/(x+3)](x+3) = 1/e10 (1)
= 1/e10 = e-10

QUESTION: 10

If a,b,c,d are positive, then

Solution: