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$L t subscript x not stretchy rightwards arrow pi over 2 end subscript left square bracket sec space 3 x cos space 5 x right square bracket$

$\frac{- 5}{3}$
$\frac{5}{3}$
$\frac{3}{5}$
$\frac{- 3}{5}$

hint Hint:

We can apply L'Hopital's rule, also commonly spelled L'Hospital's rule, whenever direct substitution of a limit yields an indeterminate form. This means that the limit of a quotient of functions (i.e., an algebraic fraction) is equal to the limit of their derivatives. Direct substitution can sometimes be used to calculate the limits for functions involving trigonometric functions.
In this question, we have to find value of $L t subscript x not stretchy rightwards arrow pi over 2 end subscript left square bracket sec space 3 x cos space 5 x right square bracket$ .

The correct answer is: $fraction numerator negative 5 over denominator 3 end fraction$

$L t subscript x not stretchy rightwards arrow pi over 2 end subscript left square bracket sec space 3 x cos space 5 x right square bracket$
Let $x space plus space h equals straight pi over 2$
$s e c space 3 x space cos 5 x$ = sec3( $straight pi over 2$ − h )cos5( $straight pi over 2$ − h )
( −cosec 3h )( sin 5h ) = $negative space fraction numerator sin space 5 h over denominator sin space 3 h end fraction$

$L t subscript h rightwards arrow 0 end subscript minus space fraction numerator sin space 5 h over denominator sin space 3 h end fraction$ = $L t subscript h not stretchy rightwards arrow 0 end subscript fraction numerator sin space 0 over denominator sin space 0 end fraction space equals space 0 over 0$
Since the limit is in the form 0 over 0, it is indeterminate—we don’t yet know what is it. We need to do some work to put it in a form where we can determine the limit.
$L t subscript h not stretchy rightwards arrow 0 end subscript fraction numerator negative sin space 5 h over denominator sin space 3 h end fraction equals space L t subscript h not stretchy rightwards arrow 0 end subscript fraction numerator negative sin space 5 h over denominator sin space 3 h end fraction cross times fraction numerator 5 cross times 3 over denominator 5 cross times 3 end fraction$
$L t subscript h not stretchy rightwards arrow 0 end subscript fraction numerator sin space 5 h over denominator 5 h end fraction cross times L t subscript h not stretchy rightwards arrow 0 end subscript fraction numerator 3 h over denominator sin space 3 h end fraction space cross times fraction numerator negative 5 over denominator 3 end fraction$
We know that, $L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin space x over denominator x end fraction space equals space 1$
$L t subscript h not stretchy rightwards arrow 0 end subscript fraction numerator sin space 5 x over denominator 5 x end fraction cross times L t subscript h not stretchy rightwards arrow 0 end subscript fraction numerator 3 x over denominator sin space 3 x end fraction space cross times fraction numerator negative 5 over denominator 3 end fraction$ = $fraction numerator negative 5 over denominator 3 end fraction$

We can only apply the L’Hospital’s rule if the direct substitution returns an indeterminate form, that means 0 over 0 or $fraction numerator plus-or-minus infinity over denominator plus-or-minus infinity end fraction.$

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The correct answer is:

Let = sec3( − h )cos5( − h )( −cosec 3h )( sin 5h ) = = Since the limit is in the form 0 over 0, it is indeterminate—we don’t yet know what is it. We need to do some work to put it in a form where we can determine the limit.We know that, =

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The correct answer is: $fraction numerator negative 5 over denominator 3 end fraction$

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