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Question

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

1
$\infty$
0
none of these

hint Hint:

In this question we have used standard limits $limit as x rightwards arrow infinity of fraction numerator sin open parentheses x close parentheses over denominator x end fraction equals 0$ and $limit as x rightwards arrow infinity of fraction numerator cos squared open parentheses x close parentheses over denominator x end fraction equals 0$ .

The correct answer is: 1

In this question we are given $f left parenthesis x right parenthesis space equals space square root of fraction numerator x minus sin open parentheses x close parentheses over denominator x plus cos squared open parentheses x close parentheses end fraction end root$ and we have to find the limit $limit as x rightwards arrow infinity of f left parenthesis x right parenthesis$
Step1: Using Standard limits
We know that $limit as x rightwards arrow infinity of fraction numerator sin open parentheses x close parentheses over denominator x end fraction equals 0$ and $limit as x rightwards arrow infinity of fraction numerator cos squared open parentheses x close parentheses over denominator x end fraction equals 0$
$limit as x rightwards arrow infinity of f left parenthesis x right parenthesis space equals space square root of fraction numerator 1 minus begin display style fraction numerator sin open parentheses x close parentheses over denominator x end fraction end style over denominator 1 plus begin display style fraction numerator cos squared open parentheses x close parentheses over denominator x end fraction end style end fraction end root$
=> $limit as x rightwards arrow infinity of f left parenthesis x right parenthesis space equals space square root of fraction numerator 1 minus begin display style 0 end style over denominator 1 plus begin display style 0 end style end fraction end root$
=> $limit as x rightwards arrow infinity of f left parenthesis x right parenthesis space equals space 1$

Related Questions to study

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parallel

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

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parallel

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parallel

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$stack a with minus on top comma stack b with minus on top comma stack c with minus on top$ 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

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