Maths-

General

Easy

Question

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis invisible function application left parenthesis e to the power of t a n invisible function application x minus e to the power of x right parenthesis divided by left parenthesis t a n invisible function application x minus x right parenthesis$

0
$e^{x}$
$\tan x$
1

hint Hint:

We are given a function. We have to find the limit of the function. We should be careful that the limit doesn't give is zero

The correct answer is: 1

The given function is
f(x) = $fraction numerator e to the power of t a n x end exponent minus e to the power of x over denominator tan x minus x end fraction$
$limit as x rightwards arrow 0 of f left parenthesis x right parenthesis space equals limit as x rightwards arrow 0 of fraction numerator e to the power of tan x end exponent minus e to the power of x over denominator tan x minus x end fraction T h e r e space i s space o n e space f o r m u l a space limit as x rightwards arrow 0 of fraction numerator e to the power of x minus 1 over denominator x end fraction equals 1$
$limit as x rightwards arrow 0 of f left parenthesis x right parenthesis space equals space limit as x rightwards arrow 0 of fraction numerator e to the power of x left parenthesis e to the power of tan x minus x end exponent minus 1 right parenthesis over denominator tan x minus x end fraction space space space space space space space space space space space space space space space equals limit as x rightwards arrow 0 of space e to the power of x space cross times space fraction numerator e to the power of tan x minus x end exponent over denominator tan x minus x end fraction space space space space space space space space space space space space space space space equals e to the power of 0 space end exponent cross times space 1 space space space space space space space space space space space space space space space equals space 1$
This is the required answer.

For such questions, we should know different formulas of limits.

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x times 2 to the power of x minus x right parenthesis divided by left parenthesis 1 minus c o s invisible function application x right parenthesis$

For such questions, we should know different formulas of limits.

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x times 2 to the power of x minus x right parenthesis divided by left parenthesis 1 minus c o s invisible function application x right parenthesis$

Maths-General

For such questions, we should know different formulas of limits.

General

Maths-

$L t left parenthesis x rightwards arrow pi divided by 2 right parenthesis left parenthesis s e c invisible function application x minus t a n invisible function application x right parenthesis divided by left parenthesis pi divided by 2 minus x right parenthesis$

Maths-General

General

Maths-

$L t subscript left parenthesis x rightwards arrow pi divided by 2 right parenthesis left square bracket s e c invisible function application 3 x c o s invisible function application 5 x right square bracket$

For such questions, we should know different formulas of limits. We should know L'Hospitals rule to solve question.

$L t subscript left parenthesis x rightwards arrow pi divided by 2 right parenthesis left square bracket s e c invisible function application 3 x c o s invisible function application 5 x right square bracket$

Maths-General

For such questions, we should know different formulas of limits. We should know L'Hospitals rule to solve question.

General

Maths-

$L t left parenthesis x rightwards arrow pi divided by 2 right parenthesis left square bracket s e c 3 x c o s 5 x right square bracket$

Maths-General

General

Chemistry-

K₁and K₂ are equilibrium constants for reaction (i) and (ii)
$N subscript 2 end subscript left parenthesis g right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis ℶ 2 N O left parenthesis g right parenthesis$ (i) $N O left parenthesis g right parenthesis 1 divided by 2 N subscript 2 end subscript left parenthesis g right parenthesis plus 1 divided by 2 O subscript 2 end subscript left parenthesis g right parenthesis minus$ (ii) Then,

Chemistry-General

General

Maths-

$L t left parenthesis x rightwards arrow 0 left parenthesis x plus 2 right parenthesis plus 4 right parenthesis left parenthesis 3 left parenthesis x plus 1 right parenthesis plus 4 right parenthesis divided by 3 left parenthesis x plus 2 right parenthesis$

Maths-General

General

Maths-

$left parenthesis L t left parenthesis x rightwards arrow 0 right parenthesis left parenthesis sin hat left parenthesis negative 1 right parenthesis x right parenthesis divided by x right parenthesis right parenthesis$

Maths-General

General

Maths-

$L t left parenthesis x rightwards arrow 0 right parenthesis 2 x divided by left parenthesis t a n negative 1 right parenthesis x right parenthesis$

Maths-General

General

Maths-

$L t left parenthesis x rightwards arrow 4 right parenthesis left parenthesis √ x minus 2 right parenthesis divided by left parenthesis x minus 4 right parenthesis$

Maths-General

General

Maths-

$L t left parenthesis x rightwards arrow 0 right parenthesis 2 divided by x l o g left parenthesis 1 plus x right parenthesis$

Maths-General

General

Maths-

$L t left parenthesis x rightwards arrow 0 right parenthesis left parenthesis 1 plus 2 x divided by 3 right parenthesis left parenthesis 1 divided by x right parenthesis$

Maths-General

General

Maths-

$Lt subscript straight x not stretchy rightwards arrow negative 1.5 end subscript COS to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 3 over denominator 3 end fraction close parentheses$

Maths-General

General

Maths-

$L t left parenthesis x rightwards arrow 2 right parenthesis l o g subscript 1 0 left square bracket x to the power of 6 plus √ left parenthesis x squared plus 1292 right parenthesis right square bracket$

Maths-General

General

Maths-

$L t left parenthesis x rightwards arrow 0 right parenthesis left parenthesis 1 minus c o s x right parenthesis divided by left parenthesis x rightwards arrow 0 right parenthesis 1 divided by x squared$

Maths-General

General

Maths-

$L t left parenthesis x rightwards arrow 0 right parenthesis left parenthesis 3 s i n x to the power of 0 minus s i n 3 x to the power of 8 right parenthesis divided by x cubed$

Maths-General

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01

We are given a function. We have to find the limit of the function. We should be careful that the limit doesn't give is zero

The correct answer is: 1

The given function isf(x) = This is the required answer.

Related Questions to study

K1 and K2 are equilibrium constants for reaction (i) and (ii) (i) (ii) Then,

K1 and K2 are equilibrium constants for reaction (i) and (ii) (i) (ii) Then,

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0
$e^{x}$
$\tan x$
1