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General

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Question

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left square bracket left parenthesis a to the power of left parenthesis 1 divided by x right parenthesis plus b to the power of left parenthesis 1 divided by x right parenthesis plus c to the power of left parenthesis 1 divided by x right parenthesis right parenthesis divided by 3 right square bracket to the power of x$ where a,b,c are real and non-zero=

$(a b c)$
$(a b c)^{(} 1 / 3)$
$(a b c)^{(} 1 / x)$
$\infty$

The correct answer is: $left parenthesis a b c right parenthesis to the power of left parenthesis 1 divided by 3 right parenthesis$

Related Questions to study

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x squared plus 5 x plus 3 right parenthesis divided by left parenthesis x squared plus x plus 2 right parenthesis right parenthesis to the power of x equals$

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x squared plus 5 x plus 3 right parenthesis divided by left parenthesis x squared plus x plus 2 right parenthesis right parenthesis to the power of x equals$

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x plus 6 right parenthesis divided by left parenthesis x plus 1 right parenthesis right parenthesis to the power of left parenthesis x plus 4 right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x plus 6 right parenthesis divided by left parenthesis x plus 1 right parenthesis right parenthesis to the power of left parenthesis x plus 4 right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x plus 5 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of left parenthesis x plus 3 right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x plus 5 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of left parenthesis x plus 3 right parenthesis equals$

parallel

$L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x plus a right parenthesis divided by left parenthesis x plus b right parenthesis right parenthesis to the power of left parenthesis x plus b right parenthesis equals$

$L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x plus a right parenthesis divided by left parenthesis x plus b right parenthesis right parenthesis to the power of left parenthesis x plus b right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis e to the power of x minus e to the power of left parenthesis i n invisible function application x right parenthesis right parenthesis divided by left parenthesis 2 left parenthesis x minus s i n invisible function application x right parenthesis right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis e to the power of x minus e to the power of left parenthesis i n invisible function application x right parenthesis right parenthesis divided by left parenthesis 2 left parenthesis x minus s i n invisible function application x right parenthesis right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis left parenthesis 1 minus e to the power of x right parenthesis s i n invisible function application x right parenthesis divided by left parenthesis x squared plus x cubed right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis left parenthesis 1 minus e to the power of x right parenthesis s i n invisible function application x right parenthesis divided by left parenthesis x squared plus x cubed right parenthesis equals$

parallel

If $a greater than 0 text and end text Lt invisible function application left parenthesis x not stretchy rightwards arrow a right parenthesis open parentheses a to the power of x minus x to the power of 4 close parentheses divided by open parentheses x to the power of x minus a cubed close parentheses equals negative 1 text , then end text a equals$

If $a greater than 0 text and end text Lt invisible function application left parenthesis x not stretchy rightwards arrow a right parenthesis open parentheses a to the power of x minus x to the power of 4 close parentheses divided by open parentheses x to the power of x minus a cubed close parentheses equals negative 1 text , then end text a equals$

If $stack a with minus on top equals left parenthesis 3 comma negative 1 , 5 right parenthesis comma stack b with minus on top equals left parenthesis 1 , 2 comma negative 3 right parenthesis$ are given vectors. A vector $stack c with minus on top$ which is perpendicular to $z$ -axis satisfying $stack c with minus on top times stack a with minus on top equals 9$ and $stack c with minus on top times stack b with minus on top equals negative 4$ then

If $stack a with minus on top equals left parenthesis 3 comma negative 1 , 5 right parenthesis comma stack b with minus on top equals left parenthesis 1 , 2 comma negative 3 right parenthesis$ are given vectors. A vector $stack c with minus on top$ which is perpendicular to $z$ -axis satisfying $stack c with minus on top times stack a with minus on top equals 9$ and $stack c with minus on top times stack b with minus on top equals negative 4$ then

$a with minus on top equals left parenthesis c o s invisible function application theta right parenthesis times i with minus on top minus left parenthesis s i n invisible function application theta right parenthesis j with minus on top comma b with minus on top equals left parenthesis s i n invisible function application theta right parenthesis times i with minus on top plus left parenthesis c o s invisible function application theta right parenthesis j with minus on top comma c with minus on top equals k with minus on top comma r with minus on top equals 17 i with minus on top plus 7 j with minus on top plus 15 k with minus on top$ and $stack r with ‾ on top equals x stack a with ‾ on top plus y stack b with ‾ on top plus z stack c with ‾ on top$ . The range of $x plus y plus z$ is

$a with minus on top equals left parenthesis c o s invisible function application theta right parenthesis times i with minus on top minus left parenthesis s i n invisible function application theta right parenthesis j with minus on top comma b with minus on top equals left parenthesis s i n invisible function application theta right parenthesis times i with minus on top plus left parenthesis c o s invisible function application theta right parenthesis j with minus on top comma c with minus on top equals k with minus on top comma r with minus on top equals 17 i with minus on top plus 7 j with minus on top plus 15 k with minus on top$ and $stack r with ‾ on top equals x stack a with ‾ on top plus y stack b with ‾ on top plus z stack c with ‾ on top$ . The range of $x plus y plus z$ is

parallel

$stack a with minus on top comma stack b with minus on top comma stack c with minus on top$ are non-coplanar vectors and $stack a with minus on top plus stack b with minus on top plus stack c with minus on top$ and $stack b with minus on top plus stack c with minus on top plus stack d with minus on top$ are collinear with $stack d with minus on top$ and $stack a with minus on top$ respectively then $stack a with minus on top plus stack b with minus on top plus stack c with minus on top plus stack d with minus on top$ is

$stack a with minus on top comma stack b with minus on top comma stack c with minus on top$ are non-coplanar vectors and $stack a with minus on top plus stack b with minus on top plus stack c with minus on top$ and $stack b with minus on top plus stack c with minus on top plus stack d with minus on top$ are collinear with $stack d with minus on top$ and $stack a with minus on top$ respectively then $stack a with minus on top plus stack b with minus on top plus stack c with minus on top plus stack d with minus on top$ is

If $a subscript 1 end subscript$ and $a subscript 2 end subscript$ are two values of a for which the unit vector $stack a i with minus on top plus b j with minus on top plus 1 half k with minus on top$ is linearly dependent with $i with minus on top plus 2 j with minus on top$ and $j with minus on top minus 2 k with minus on top$ then $fraction numerator 1 over denominator a subscript 1 end subscript end fraction plus fraction numerator 1 over denominator a subscript 2 end subscript end fraction$ is equal to

If $a subscript 1 end subscript$ and $a subscript 2 end subscript$ are two values of a for which the unit vector $stack a i with minus on top plus b j with minus on top plus 1 half k with minus on top$ is linearly dependent with $i with minus on top plus 2 j with minus on top$ and $j with minus on top minus 2 k with minus on top$ then $fraction numerator 1 over denominator a subscript 1 end subscript end fraction plus fraction numerator 1 over denominator a subscript 2 end subscript end fraction$ is equal to

$a with rightwards arrow on top equals left parenthesis c o s invisible function application theta right parenthesis i with rightwards arrow on top minus left parenthesis s i n invisible function application theta right parenthesis j with rightwards arrow on top comma b with rightwards arrow on top equals left parenthesis s i n invisible function application theta right parenthesis i with rightwards arrow on top plus left parenthesis c o s invisible function application theta right parenthesis j with rightwards arrow on top comma c with rightwards arrow on top equals k with rightwards arrow on top comma r with rightwards arrow on top equals 7 i with rightwards arrow on top plus j with rightwards arrow on top plus 10 k with rightwards arrow on top$ if $stack r with rightwards arrow on top equals x stack a with rightwards arrow on top plus y stack b with rightwards arrow on top plus z stack c with rightwards arrow on top$ , then

$a with rightwards arrow on top equals left parenthesis c o s invisible function application theta right parenthesis i with rightwards arrow on top minus left parenthesis s i n invisible function application theta right parenthesis j with rightwards arrow on top comma b with rightwards arrow on top equals left parenthesis s i n invisible function application theta right parenthesis i with rightwards arrow on top plus left parenthesis c o s invisible function application theta right parenthesis j with rightwards arrow on top comma c with rightwards arrow on top equals k with rightwards arrow on top comma r with rightwards arrow on top equals 7 i with rightwards arrow on top plus j with rightwards arrow on top plus 10 k with rightwards arrow on top$ if $stack r with rightwards arrow on top equals x stack a with rightwards arrow on top plus y stack b with rightwards arrow on top plus z stack c with rightwards arrow on top$ , then

parallel

$L t ┬ left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x left parenthesis 1 plus a c o s invisible function application x right parenthesis minus b s i n invisible function application x right parenthesis divided by x cubed equals 1 t h e n a equals comma b equals$

$L t ┬ left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x left parenthesis 1 plus a c o s invisible function application x right parenthesis minus b s i n invisible function application x right parenthesis divided by x cubed equals 1 t h e n a equals comma b equals$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x 10 to the power of x minus x right parenthesis divided by left parenthesis 1 minus c o s invisible function application x right parenthesis equals$

For such questions, we should know different formulae.

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x 10 to the power of x minus x right parenthesis divided by left parenthesis 1 minus c o s invisible function application x right parenthesis equals$

For such questions, we should know different formulae.

$Cu to the power of plus 2 end exponent$ ions are more stable than $Cu to the power of plus$ ions because ions _________ as compared to $Cu to the power of plus 2 end exponent$ ions

$Cu to the power of plus 2 end exponent$ ions are more stable than $Cu to the power of plus$ ions because ions _________ as compared to $Cu to the power of plus 2 end exponent$ ions

Chemistry-General

parallel

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