límits i continuïtat de funcions - matessantboianes

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⎛ 2 2 x x a) lim ⎜ − 1 1+ 2 ⎞ ⎜ − → ` − ` x→ + `⎝⎜ x 2x−1 ⎠⎟ x ` x ` x→ + ` 2 −x −3x + 1 1 =− 2 2 x − x 2 Solucionari ⎛ x − + x ⎞ lim ⎜ − lim →+ ⎝⎜ x x − ⎠⎟ →+ = 2 2 2 2 1 1 2 ( x −1)( 2x −1) − ( 1+ 2 x ) x = 2 1 x( 2x − 1) b) lim x→ + ` = lim ( 2 x −2x − x )= lim x→+ ` x→+ ` c) lim x→ + ` lim ( 2 x −2x − x ) → ` − ` = lim x→ + ` ( 2x − 1+ 4 x ) → ` − ` lim ( 2x − 1+ 4 x )= lim x→+ ` x→ + ` d) lim x→ + ` lim x→ + ` ( 2 9x + 3x −3x) → ` − ` ( 9x + 3x −3x)= 2 lim x→ + ` = lim x→ + ` 2 2 x −2x − x = 2 x − 2 x + x −2 x =−1 2 x − 2 x + x 2 4 x −1−4x 2x + 1+ 4 x 020 Substitueix a, b, c i d per nombres de manera que: a) lim x → + ` b) lim x → + ` ( x + ax − x )= 2 1 ( 2 4 x + bx−3 − 2 x )=− 1 4 ( ) = =+ ` 2 2 9x + 3x −9x = 2 9x + 3x + 3 x 3 x 1 = 2 9x+ 3x + 3 x 2 c) lim x → + ` d) lim x → + ` 2 2 ( 2 9x + 7 −cx)= 0 7 ( dx + x− − x )=+ 2 5 2 ` a) lim 2 x + ax − x lim x + ax − x = lim ax = x→+ ` x→+ ` 2 x + ax + x x→ + ` 2 x + ax + x = a 2 = 1→a= 2 b) lim ( 2 4 x + bx −3 − 2 x)= lim x→+ ` x→+ ` = lim x→ + ` 2 2 4 x + bx −3−4x = 2 4 x + bx − 3 + 2x bx − 3 b 1 = =−→ b =−1 2 4 x + bx − 3 + 2 x 4 4 419
420 límits i continuïtat de funcions c) lim ( 2 9x + 7 − cx)= lim x→+ ` x→+ ` d) lim ( 2 dx + x −5 − 2x)= lim x→+ ` x→+ ` 021 calcula els límits següents: a) lim x → + ` b) lim x → + ` ⎛ ⎜ 1 ⎞ ⎜ 1+ ⎟ ⎝⎜ ⎟ x ⎠⎟ ⎛ ⎜ 3 ⎞ ⎜ 1− ⎟ ⎝⎜ ⎟ x ⎠⎟ a) lim x→ + ` lim x→ + ` b) lim x→ + ` lim x→ + ` 2 x−1 6 x+ 2 ⎛ ⎜ 1 ⎞ ⎜1+ ⎝⎜ x ⎠⎟ ⎛ 1 ⎞ ⎜ ⎜1+ ⎝⎜ x ⎠⎟ ⎛ ⎜ 3 ⎞ ⎜1− ⎝⎜ x ⎠⎟ ⎛ 3 ⎞ ⎜ ⎜1− ⎝⎜ x ⎠⎟ 2 x−1 → 1 ` 2 2 2 9x + 7− c x c) lim 2 cx 9x+ 7 + x →−` ⎡ ⎤ 2 x−1 ⎢ 1 lim ⋅( 2 x−1) ⎥ ⎢ x ⎥ 6 x+ 2 ⎛ x c) lim ⎜ 4 − 1⎞ ⎜ 2 − x→ −`⎝⎜ 4 x ⎠⎟ ⎛ lim ⎜ 4 x − 1⎞ ⎜ 2 − → −`⎝⎜ 4 x ⎠⎟ x ⎛ x ⎞ d) lim ⎜ x→ −`⎝⎜ x + 3 ⎠⎟ ⎛ x ⎞ lim ⎜ → −`⎝⎜ x + 3 ⎠⎟ 022 Troba aquests límits: a) lim x x → + ` b) lim x → + ` ⎛ 2 ⎜ x + 1 ⎞ ⎜ ⎟ ⎝⎜ 2 x ⎠⎟ x−1 2 2 = 0 → c = 3 dx + x −5−4 x =+ ` → d ≠ 4 2 dx + x − 5 + 2x ⎛ ⎜ 4 x −1 ⎞ ⎜ 2 − ⎟ ⎝⎜ ⎟ 4 x ⎠⎟ ⎛ x ⎞ d) lim ⎜ ⎟ x →− ⎝⎜ ⎟ ` x + 3 ⎠⎟ x→ + ` = e ⎣ ⎦ = e → 1 ` + − = ⎛ ⎞ ⎝⎜ ⎠⎟ ⋅ 6 x 2 3 6 + x e x ⎡ ⎤ lim ⎢ ( x+ → ` ⎢ 2 ) ⎥ ⎣ ⎦ −18 e 3 x+ 1 3 + 1 x ⎛ 2 ⎜ x − x + 2 ⎞ ⎜ ⎟ ⎝⎜ 2 x + 1 ⎠⎟ x+ 6 3 x+ 2 3 x+ 2 → 1 = ` → 1 = e ` 2 = = ⎡ ⎛ ⎜ 4x−1⎞ ⎤ lim ⎢ 1− x→ −` ⎢ ⋅ ( 3 x+ 2) ⎥ ⎜⎝ 4 x ⎠⎟ ⎥ ⎣ ⎦ + − ⎛ ⎞ + ⎝⎜ 3 ⎠⎟ 1 x x e x ⎡ ⎤ lim ⎢ ⋅ + → ` ⎢ ( 3 x 1) ⎥ ⎣ ⎦ c) lim x →−` d) lim x →−` ⎛ ⎜ x ⎜ ⎝⎜ x 3 x+ 1 e 3 x+ 2 1 18 3x+ 2 lim x→ − = e ` 4 x = e 3 4 4 3 = e − 3( 3 x+ 1) lim x→ −` x+ 3 −9 1 9 = e = e = e 2 2 −1 ⎞ ⎟ + 1 ⎠⎟ 1 2 x ⎛ 2 ⎜ x − 2x + 2 ⎞ ⎜ ⎟ ⎝⎜ 2 2x + 3x−2⎠⎟ 2 x −3 x
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