Foundations/Comprehension 1.The following limits are of the form 0/0. Using algebraic methods, simplify the expressions, then compute the limits (5 points each). a)limx→103x2−23x−7011x2−1100 b)limh→05(8+h)2−320h

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Foundations/Comprehension    1.The following limits are of the form 0/0. Using algebraic methods, simplify the expressions, then compute the limits (5 points each).    a)limx→103x2−23x−7011x2−1100    b)limh→05(8+h)2−320h

Arnold Odonnell · Accepted Answer

Step 1  a)  limx→103x2−23x−7011x2−1100  As x→10 then numerator and denominator both tends to zero.  i⋅e limit is of the form 00 form.  Now limx→103x2−23x−7011x2−1100  =limx→103x2−30x+7x−7011x2−1100  =limx→103x(x−10)+7(x−10)11(x2−100) (factoring numerator and denominator)  =limx→10(3x+7)(x−10)11(x−10)(x+10) (∵a2−b2=(a−b)(a+b))  =limx→103x+711(x+10) (concelling the common factor)  =3(10)+711(10+10) (finally compuling the limit)  =3(10)+711(+10)  Final fons: limx→103x2−23x−7011x2−1100=37220 Ans.  b) limh→05(8+h)2−320h  As h→0, both numerator and denominator tends to zero, so limit is of  00 form.  Now limh→05(8+h)2−320h  limh→05(82+h2+16h)−320h  limh→0¬{320}+5h2+16h−¬{320}h (simplifying the numerator)

Foundations/Comprehension 1.The following limits are of the form 0/0. Using algebraic

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