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Question

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x plus 6 right parenthesis divided by left parenthesis x plus 1 right parenthesis right parenthesis to the power of left parenthesis x plus 4 right parenthesis equals$

$e^{6}$
$e^{4}$
e
$e^{3}$

hint Hint:

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

The correct answer is: $e cubed$

In this question we have to find the limit of $limit as x rightwards arrow infinity of open parentheses fraction numerator x plus 6 over denominator x plus 1 end fraction close parentheses to the power of left parenthesis x plus 4 right parenthesis end exponent$
Step1: Putting the value of limit in the expression.
By putting the value of limit we are get $1 to the power of infinity$ form.
Step2: Using Standard limits
We know that $limit as x rightwards arrow infinity of left parenthesis f left parenthesis x right parenthesis to the power of g left parenthesis x right parenthesis end exponent equals space e to the power of limit as x rightwards arrow infinity of left parenthesis f left parenthesis x right parenthesis minus 1 right parenthesis g left parenthesis x right parenthesis end exponent$
=> $e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator x plus 6 over denominator x plus 1 end fraction minus 1 close parentheses left parenthesis x plus 4 right parenthesis end exponent$
=> $e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator x plus 6 minus x minus 1 over denominator x plus 1 end fraction close parentheses left parenthesis x plus 4 right parenthesis end exponent$
=> $e to the power of limit as x rightwards arrow infinity of 5 cross times open parentheses fraction numerator x plus 4 over denominator x plus 1 end fraction close parentheses end exponent$
=> $e to the power of limit as x rightwards arrow infinity of 5 cross times open parentheses fraction numerator 1 plus begin display style 4 over x end style over denominator 1 plus begin display style 1 over x end style end fraction close parentheses end exponent$
Step3: Putting the value of limit
=> $e to the power of 5$
Hence, the value of limit is $e to the power of 5$ .

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

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

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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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For such questions, we should know different formulae.

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For such questions, we should know different formulae.

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