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Maths-

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Question

$L t subscript left parenthesis x rightwards arrow negative infinity right parenthesis left square bracket left parenthesis x to the power of 4 s i n invisible function application left parenthesis 1 divided by x right parenthesis plus x squared right parenthesis divided by left parenthesis left parenthesis 1 plus vertical line x vertical line cubed right parenthesis right parenthesis right square bracket equals$

1
-1
0
$- \infty$

hint Hint:

In this question first we will rearrange the expression by changing $open vertical bar x close vertical bar equals negative x$ as $x$ will always be negative. Then we divide both denominator and numerator by $x cubed$ . and then we use standard limits $limit as x minus greater than 0 of fraction numerator sin open parentheses theta close parentheses over denominator theta end fraction equals 1$ .

The correct answer is: -1

In this question we have to find the limit of $limit as x rightwards arrow negative infinity of open parentheses fraction numerator x to the power of 4 sin open parentheses begin display style 1 over x end style close parentheses plus x squared over denominator 1 plus open vertical bar x close vertical bar cubed end fraction close parentheses$
Step1: Rearranging the expression.
Since $x minus greater than negative infinity$ , $x$ will always be negative, $open vertical bar x close vertical bar equals negative x$
$limit as x rightwards arrow negative infinity of open parentheses fraction numerator x to the power of 4 sin open parentheses begin display style 1 over x end style close parentheses plus x squared over denominator 1 minus x cubed end fraction close parentheses$

By dividing both numerator and denominator by $x cubed$ , we get
=> $limit as x rightwards arrow negative infinity of open parentheses fraction numerator begin display style fraction numerator sin open parentheses begin display style 1 over x end style close parentheses over denominator begin display style 1 over x end style end fraction end style plus begin display style 1 over x end style over denominator begin display style 1 over x cubed end style minus 1 end fraction close parentheses$
Step2: Using standard limits
We know that $limit as x minus greater than 0 of fraction numerator sin open parentheses theta close parentheses over denominator theta end fraction equals 1$
=> $limit as x rightwards arrow negative infinity of open parentheses fraction numerator begin display style 1 end style plus begin display style 1 over x end style over denominator begin display style 1 over x cubed end style minus 1 end fraction close parentheses$
By putting the value of limit we get
=> $fraction numerator 1 plus 0 over denominator 0 minus 1 end fraction equals negative 1$
Hence, the value of the limit is $negative 1$ .

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

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

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.

parallel

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