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Mademuasel [1]
4 years ago
6

Find the given limit lim(x,y)→(a,a)x4−y4x2−y2=lim(x,y)→(a,a) (x^4−y^4)/(x2−y2)= here aa is a constant. (find the limit assuming

x≠y)
Mathematics
1 answer:
Aliun [14]4 years ago
7 0
If x\neq y, then we can write

\dfrac{x^4-y^4}{x^2-y^2}=\dfrac{(x^2-y^2)(x^2+y^2)}{x^2-y^2}=x^2+y^2

and since this is continuous for all x\neq y, the limit as (x,y)\to(a,a) will be

\displaystyle\lim_{(x,y)\to(a,a)}\frac{x^4-y^4}{x^2-y^2}=\lim_{(x,y)\to(a,a)}(x^2+y^2)=2a^2
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A jar contains 12 red marbles numbered 1 to 12 and 6 blue marbles numbered 1 to 6. A marble is drawn at random from the jar. Fin
lora16 [44]

Answer:

(a)P(Red)=\dfrac{2}{3}\\(b)P(Odd) =\dfrac{1}{2}\\(c)P(\text{Red or Odd Numbered})=\dfrac{5}{6}\\(d)P(\text{Blue or Even Numbered})=\dfrac{2}{3}

Step-by-step explanation:

Number of Red Marbles{1,2,3,4,5,6,7,8,9,10,11,12},n(R)=12

Number of Blue Marbles{1,2,3,4,5,6},n(B)=6

Total Number of Marbles, n(S)=6+12=18

(a)Probability that the Marble is Red

P(R)=\dfrac{n(R)}{n(S)} =\dfrac{12}{18} =\dfrac{2}{3}

(b)Probability that the marble is odd-numbered.

Number of Odd-Numbered Balls, n(O)=9

P(Odd)=\dfrac{n(O)}{n(S)} =\dfrac{9}{18} =\dfrac{1}{2}

(c)Probability that the marble is red or odd-numbered.

n(Red)=12

n(Odd Numbered marbles)=9

n(Red and Odd Numbered Marbles)=6

P(\text{Red or Odd Numbered})=P(Red)+P(Odd\:Numbered)-P(\text{Red and Odd Numbered)}\\=\dfrac{12}{18} +\dfrac{9}{18}-\dfrac{6}{18}  =\dfrac{15}{18}\\P(\text{Red or Odd Numbered})=\dfrac{5}{6}(d)Probability that the marble is blue or even-numbered.

n(Blue)=6

n(Even Numbered marbles)=9

n(Blue and Even Numbered Marbles)=3

P(\text{Blue or Even Numbered})=P(Blue)+P(Even\:Numbered)-P(\text{Blue and Even Numbered)}\\=\dfrac{6}{18} +\dfrac{9}{18}-\dfrac{3}{18}  =\dfrac{12}{18}\\P(\text{Blue or Even Numbered})=\dfrac{2}{3}

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Step-by-step explanation:

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