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$Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 2 x over denominator tan to the power of negative 1 end exponent space x end fraction$

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hint Hint:

We can apply L'Hopital's rule, also commonly spelled L'Hospital's rule, whenever direct substitution of a limit yields an indeterminate form. This means that the limit of a quotient of functions (i.e., an algebraic fraction) is equal to the limit of their derivatives.
In this question, we have to find value of $Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 2 x over denominator tan to the power of negative 1 end exponent space x end fraction$ .

The correct answer is: 2

$Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 2 x over denominator tan to the power of negative 1 end exponent space x end fraction$
We first try substitution:
$Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 2 x over denominator tan to the power of negative 1 end exponent space x end fraction$ = $Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 0 over denominator tan to the power of negative 1 end exponent space 0 end fraction space equals 0 over 0$
Since the limit is in the form 0 over 0, it is indeterminate—we don’t yet know what is it. We need to do some work to put it in a form where we can determine the limit.
$Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 2 x over denominator tan to the power of negative 1 end exponent space x end fraction$ ( We know that $Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator tan to the power of negative 1 end exponent space x over denominator x end fraction space equals space 1$ )
On substituting, We get
$Lt subscript x not stretchy rightwards arrow 0 end subscript 2$ = 2
Also, We can solve using L'Hospital's rule ; $Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction space equals space fraction numerator f apostrophe left parenthesis 0 right parenthesis over denominator g apostrophe left parenthesis 0 right parenthesis end fraction$ .

We can only apply the L’Hospital’s rule if the direct substitution returns an indeterminate form, that means $fraction numerator 0 over denominator 0 space end fraction space o r space fraction numerator plus-or-minus infinity over denominator plus-or-minus infinity end fraction.$

One mole of an ideal gas $left parenthesis gamma equals 7 divided by 5 right parenthesis$ is adiabatically compressed so that its temperature rises from $27 to the power of 0 end exponent C text to end text 35 to the power of 0 end exponent C$ The work done by the gas is (R=8.47J/mol/K)

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$Lt subscript x not stretchy rightwards arrow 4 end subscript fraction numerator square root of x minus 2 over denominator x minus 4 end fraction$

We can only apply the L’Hospital’s rule if the direct substitution returns an indeterminate form, that means $0 over 0 space o r space fraction numerator plus-or-minus infinity over denominator plus-or-minus infinity end fraction.$

$Lt subscript x not stretchy rightwards arrow 4 end subscript fraction numerator square root of x minus 2 over denominator x minus 4 end fraction$

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If $0 less than alpha comma beta less than fraction numerator pi over denominator 4 end fraction comma C o s invisible function application left parenthesis alpha plus beta right parenthesis equals fraction numerator 4 over denominator 5 end fraction comma S i n invisible function application left parenthesis alpha minus beta right parenthesis equals fraction numerator 5 over denominator 13 end fraction$ then $Tan invisible function application 2 alpha$ =

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$Tan invisible function application 20 to the power of ring operator plus 2 Tan invisible function application 50 to the power of ring operator equals$

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$Tan invisible function application 20 to the power of ring operator plus Tan invisible function application 25 to the power of ring operator plus Tan invisible function application 20 to the power of ring operator Tan invisible function application 25 to the power of ring operator equals$

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If $Tan invisible function application 22 to the power of ring operator plus Tan invisible function application 38 to the power of ring operator minus square root of 3 equals k Tan invisible function application 22 to the power of ring operator Tan invisible function application 38 to the power of ring operator$ then k=

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$Lt subscript x not stretchy rightwards arrow 0 end subscript space open parentheses 1 plus fraction numerator 2 x over denominator 3 end fraction close parentheses to the power of 1 over x end exponent$

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Which one of the following quantities does not have unit of force per unit area.

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If $f left parenthesis x right parenthesis equals open curly brackets table row cell x to the power of alpha end exponent l o g invisible function application x comma end cell cell x greater than 0 end cell row cell 0 comma end cell cell x equals 0 end cell end table close$ and Rolle's theorem is applicable to f(x) for $x element of left square bracket 0 , 1 right square bracket$ then $alpha$ is equal to

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The correct answer is: 2

We first try substitution: = Since the limit is in the form 0 over 0, it is indeterminate—we don’t yet know what is it. We need to do some work to put it in a form where we can determine the limit. ( We know that )On substituting, We get = 2Also, We can solve using L'Hospital's rule ; .

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One mole of an ideal gas $left parenthesis gamma equals 7 divided by 5 right parenthesis$ is adiabatically compressed so that its temperature rises from $27 to the power of 0 end exponent C text to end text 35 to the power of 0 end exponent C$ The work done by the gas is (R=8.47J/mol/K)

One mole of an ideal gas $left parenthesis gamma equals 7 divided by 5 right parenthesis$ is adiabatically compressed so that its temperature rises from $27 to the power of 0 end exponent C text to end text 35 to the power of 0 end exponent C$ The work done by the gas is (R=8.47J/mol/K)

If $Tan invisible function application 40 to the power of ring operator plus 2 Tan invisible function application 10 to the power of ring operator equals Cot invisible function application x$ then x=

If $Tan invisible function application 40 to the power of ring operator plus 2 Tan invisible function application 10 to the power of ring operator equals Cot invisible function application x$ then x=

If A, B, C are acute angles, $T a n invisible function application A equals 1 divided by 2$ , $T a n invisible function application B equals 1 divided by 5$ , $Tan invisible function application C equals 1 divided by 8$ then A+B+C=

If A, B, C are acute angles, $T a n invisible function application A equals 1 divided by 2$ , $T a n invisible function application B equals 1 divided by 5$ , $Tan invisible function application C equals 1 divided by 8$ then A+B+C=