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Question

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

1
2
-1
$1 / 2$

hint Hint:

In this question we will rearrange the expression and then we will use the standard limit. $limit as x minus greater than 0 of fraction numerator e to the power of x minus 1 over denominator x end fraction equals 1$

The correct answer is: $space 1 divided by 2$

In this question we have to find the limit of $limit as x minus greater than 0 of fraction numerator e to the power of x minus e to the power of sin open parentheses x close parentheses end exponent over denominator 2 left parenthesis x minus sin open parentheses x close parentheses right parenthesis end fraction$
Step1: Putting the value of limit.
By putting the value of limit we are getting both numerator and denominator as zero. That is $0 over 0$ form.
Step2: Rearranging the expression

We will rearrange the expression by taking $e to the power of sin open parentheses x close parentheses end exponent$ common.
=> $limit as x minus greater than 0 of fraction numerator e to the power of sin open parentheses x close parentheses end exponent left parenthesis e to the power of x minus sin open parentheses x close parentheses end exponent minus 1 right parenthesis over denominator 2 left parenthesis x minus sin open parentheses x close parentheses right parenthesis end fraction$
=> $limit as x minus greater than 0 of e to the power of sin open parentheses x close parentheses end exponent over 2 cross times limit as x minus greater than 0 of fraction numerator left parenthesis e to the power of x minus sin open parentheses x close parentheses end exponent minus 1 right parenthesis over denominator left parenthesis x minus sin open parentheses x close parentheses right parenthesis end fraction$
Step3: Using the standard limits
We know that $limit as x minus greater than 0 of fraction numerator e to the power of x minus 1 over denominator x end fraction equals 1$
=> $limit as x minus greater than 0 of e to the power of sin open parentheses x close parentheses end exponent over 2 cross times limit as x minus greater than 0 of fraction numerator left parenthesis e to the power of x minus sin open parentheses x close parentheses end exponent minus 1 right parenthesis over denominator left parenthesis x minus sin open parentheses x close parentheses right parenthesis end fraction$

$limit as x minus greater than 0 of e to the power of sin open parentheses x close parentheses end exponent over 2$
Now putting the value of limit we get
= $1 half$
Hence, the value of the limit is $1 half$

Related Questions to study

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

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

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

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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

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$a with minus on top equals left parenthesis c o s invisible function application theta right parenthesis times i with minus on top minus left parenthesis s i n invisible function application theta right parenthesis j with minus on top comma b with minus on top equals left parenthesis s i n invisible function application theta right parenthesis times i with minus on top plus left parenthesis c o s invisible function application theta right parenthesis j with minus on top comma c with minus on top equals k with minus on top comma r with minus on top equals 17 i with minus on top plus 7 j with minus on top plus 15 k with minus on top$ 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

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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

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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.

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parallel

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Chemistry-General

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