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Question

$L i m left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x t a n 2 x minus 2 x t a n x right parenthesis divided by left parenthesis left parenthesis 1 minus c o s invisible function application 2 x right parenthesis squared right parenthesis equals$

2
-2
$1 / 2$
$- 1 / 2$

hint Hint:

In this question first we have to use trigonometric formula $tan open parentheses 2 x close parentheses equals fraction numerator 2 tan open parentheses x close parentheses over denominator 1 minus tan squared open parentheses x close parentheses end fraction$ and $cos open parentheses 2 x close parentheses equals fraction numerator 1 minus tan squared open parentheses x close parentheses over denominator 1 plus tan squared open parentheses x close parentheses end fraction$ and then we will use standard limits $limit as x minus greater than 0 of fraction numerator x over denominator tan open parentheses x close parentheses end fraction equals 1$

The correct answer is: $1 divided by 2$

In this question we have to find the limit $limit as x minus greater than 0 of fraction numerator x tan open parentheses 2 x close parentheses minus 2 x tan open parentheses x close parentheses over denominator left parenthesis 1 minus cos open parentheses 2 x close parentheses right parenthesis squared end fraction$
Step1: Using trigonometric formulae.
we know that $tan open parentheses 2 x close parentheses equals fraction numerator 2 tan open parentheses x close parentheses over denominator 1 minus tan squared open parentheses x close parentheses end fraction$ and $cos open parentheses 2 x close parentheses equals fraction numerator 1 minus tan squared open parentheses x close parentheses over denominator 1 plus tan squared open parentheses x close parentheses end fraction$
=> $limit as x minus greater than 0 of fraction numerator x begin display style fraction numerator 2 tan open parentheses x close parentheses over denominator 1 minus tan squared open parentheses x close parentheses end fraction end style minus 2 x tan open parentheses x close parentheses over denominator open parentheses 1 minus begin display style fraction numerator 1 minus tan squared open parentheses x close parentheses over denominator 1 plus tan squared open parentheses x close parentheses end fraction end style close parentheses squared end fraction$
=> $limit as x minus greater than 0 of fraction numerator 2 x tan open parentheses x close parentheses minus 2 x tan open parentheses x close parentheses left parenthesis 1 minus tan squared open parentheses x close parentheses right parenthesis over denominator open parentheses begin display style fraction numerator 1 plus tan squared open parentheses x close parentheses minus 1 plus tan squared open parentheses x close parentheses over denominator 1 plus tan squared open parentheses x close parentheses end fraction end style close parentheses squared end fraction$
=> $limit as x minus greater than 0 of fraction numerator 2 x tan cubed open parentheses x close parentheses over denominator 4 tan to the power of 4 open parentheses x close parentheses end fraction$
Step2: By using Standard limits
We know that $limit as x minus greater than 0 of fraction numerator x over denominator tan open parentheses x close parentheses end fraction equals 1$
=> $limit as x minus greater than 0 of fraction numerator x over denominator 2 tan open parentheses x close parentheses end fraction$
=> $1 half$ .
Hence, the value of limit is $1 half$

Related Questions to study

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If $f left parenthesis x right parenthesis equals √ left parenthesis left parenthesis x minus sin invisible function application x right parenthesis divided by left parenthesis x plus cos invisible function application x right parenthesis right parenthesis$ then $L t f left parenthesis x right parenthesis space$ is

parallel

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

parallel

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

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parallel

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parallel

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

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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

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