lim(x- -3) (x^2 - 9)/(root[x^2 + 16]-5)

that's limit(x tends to -3)

wow nice question buddy.... hold a sec ill try

Ok its an easy one... heres ur answer....

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= limit (x tending to -3) of { [x^2 - 9] / [root(x^2 + 16) - 5] }

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Multiplying and dividing by[root(x^2 + 16) + 5] to rationalise the denominator:-

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=limit (x tending to -3) of {[x^2 - 9] *[root(x^2 + 16) + 5]/ [root(x^2 + 16) - 5] *[root(x^2 + 16) + 5]}

=limit (x tending to -3) of { [x^2 - 9] *[root(x^2 + 16) + 5]/ [x^2 + 16 - 5²] }

=limit (x tending to -3) of { [x^2 - 9] *[root(x^2 + 16) + 5]/ [x^2 + 16 - 25] }

= limit(x tending to -3) of { [x^2 - 9] *[root(x^2 + 16) + 5]/ [x^2 - 9] }

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Here we can cancel out[x^2 - 9] in denominator and numerator.

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=limit(x tending to -3) of { [x^2 - 9] *[root(x^2 + 16) + 5]/ [x^2 - 9] }

= limit(x tending to -3) of [root(x^2 + 16) + 5]

= [root(9+16) + 5]

= root(25) + 5

= 5 + 5

= 10

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