Solution for the question - (4x rarr3)(x^(3)-8x^(2)+45)//(2x^(2)-3x-9) (-1)//3 '/> 1//3 '/> (-7)//3 '/> 7//3 '/>

Maths-

General

Easy

Question

$L t subscript left parenthesis x rightwards arrow 3 right parenthesis invisible function application left parenthesis x cubed minus 8 x squared plus 45 right parenthesis divided by left parenthesis 2 x squared minus 3 x minus 9 right parenthesis$

$7 / 3$
$(- 7) / 3$
$1 / 3$
$(- 1) / 3$

hint Hint:

In this question we are getting $0 over 0$ form. So, we will differentiate both numerator and denominator to find the limit.

The correct answer is: $left parenthesis negative 7 right parenthesis divided by 3$

In this given question we have to find limit of $limit as x minus greater than 3 of fraction numerator x cubed minus 8 x squared plus 45 over denominator 2 x squared minus 3 x minus 9 end fraction$
Step1: Finding the limit.
The limit can be found out by differentiating both numerator and denominator as we are getting $0 over 0$ form.
=> $limit as x minus greater than 3 of fraction numerator 3 x squared minus 16 x over denominator 4 x minus 3 end fraction$
Now we will put the value of $x equals 3$ in both numerator and denominator to find the limit.
=> $limit as x minus greater than 3 of fraction numerator 3 cross times 3 squared minus 16 cross times 3 over denominator 4 cross times 3 minus 3 end fraction$
=> $fraction numerator negative 21 over denominator 9 end fraction$
=> $fraction numerator negative 7 over denominator 3 end fraction$ .
So, the value of the limit is $fraction numerator negative 7 over denominator 3 end fraction$

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

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

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

Maths-General

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

General

Maths-

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

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

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In this question we are getting form. So, we will differentiate both numerator and denominator to find the limit.

The correct answer is:

In this given question we have to find limit of Step1: Finding the limit.The limit can be found out by differentiating both numerator and denominator as we are getting form.=> Now we will put the value of in both numerator and denominator to find the limit.=> => =>.So, the value of the limit is

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In this question we are getting $0 over 0$ form. So, we will differentiate both numerator and denominator to find the limit.

The correct answer is: $left parenthesis negative 7 right parenthesis divided by 3$