Solution for the question - for x in r(t(x rarr oo)((x-3)//(x+2))^(x)=:} e5e-5e-1 '/>e

Maths-

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Question

For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$

e
$e - 1$
e^-5
e⁵

hint Hint:

In this question we are getting $1 to the power of infinity$ form. We will use Standard limits formula $limit as x rightwards arrow infinity of f left parenthesis x right parenthesis to the power of g left parenthesis x right parenthesis end exponent equals space e to the power of limit as x rightwards arrow infinity of left parenthesis f left parenthesis x right parenthesis minus 1 right parenthesis g left parenthesis x right parenthesis end exponent$ to find the limit

The correct answer is: e^-5

In this question we have to find the limit of $limit as x rightwards arrow infinity of open parentheses fraction numerator x minus 3 over denominator x plus 2 end fraction close parentheses to the power of x$
Step1: Rearranging the equation
By dividing by $x$ in both numerator and denominator we get,
$limit as x rightwards arrow infinity of open parentheses fraction numerator 1 minus begin display style 3 over x end style over denominator 1 plus begin display style 2 over x end style end fraction close parentheses to the power of x$
By putting the value of limit we are getting $1 to the power of infinity$ form.
Step2: Using Standard limits
We know that $limit as x rightwards arrow infinity of f left parenthesis x right parenthesis to the power of g left parenthesis x right parenthesis end exponent equals space e to the power of limit as x rightwards arrow infinity of left parenthesis f left parenthesis x right parenthesis minus 1 right parenthesis g left parenthesis x right parenthesis end exponent$
$e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator 1 minus begin display style 3 over x end style over denominator 1 plus begin display style 2 over x end style end fraction minus 1 close parentheses x end exponent$
=> $e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator 1 minus begin display style 3 over x minus 1 minus 2 over x end style over denominator 1 plus begin display style 2 over x end style end fraction close parentheses x end exponent$
=> $e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator negative begin display style 5 over x end style over denominator begin display style fraction numerator 2 plus x over denominator x end fraction end style end fraction close parentheses x end exponent$
=> $e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator negative begin display style 5 x end style over denominator begin display style x plus 2 end style end fraction close parentheses end exponent$
=> $e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator negative begin display style 5 end style over denominator begin display style 1 plus fraction numerator begin display style 2 end style over denominator x end fraction end style end fraction close parentheses end exponent$
By putting the value of limits we get
$e to the power of negative 5 end exponent$
Hence, the value of limit us $e to the power of negative 5 end exponent$ .

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For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$

e
$e - 1$
e^-5
e⁵

The correct answer is: e^-5

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For

ee-5e5

In this question we are getting form. We will use Standard limits formula to find the limit

The correct answer is: e-5

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For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$

e
$e - 1$
e^-5
e⁵

The correct answer is: e^-5