Jawab:
Penjelasan dengan langkah-langkah:
limit (x->0) 3x / {√(9+x) - √(9-x)} =
limit (x->0) (3x){√(9+x)+√(9-x)} / {√(9+x)-√(9+x)}{√(9+x)+√(9-x)}
limit (x->0) (3x){√(9+x) +√(9-x)}/ ((9+x) -(9-x))
limit (x->0) {3x)√(9+x) +√(9-x)} / (2x)
limit (x->0) (3/2) { √(9+x) +√(9-x)}
x = 0 --> limit = 3/2 (√9 +√9) = 3/2 (3+3) = 3/2(6) = 9
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Jawab:
Penjelasan dengan langkah-langkah:
Jawab:
Penjelasan dengan langkah-langkah:
limit (x->0) 3x / {√(9+x) - √(9-x)} =
limit (x->0) (3x){√(9+x)+√(9-x)} / {√(9+x)-√(9+x)}{√(9+x)+√(9-x)}
limit (x->0) (3x){√(9+x) +√(9-x)}/ ((9+x) -(9-x))
limit (x->0) {3x)√(9+x) +√(9-x)} / (2x)
limit (x->0) (3/2) { √(9+x) +√(9-x)}
x = 0 --> limit = 3/2 (√9 +√9) = 3/2 (3+3) = 3/2(6) = 9