Math  /  Calculus

Question6.0 Sulla base delle informazioni che puoi dedurre dal grafico, completa le uguaglianze seguenti. a. limxf(x)=\lim _{x \rightarrow-\infty} f(x)= \qquad f. limx1f(x)=\lim _{x \rightarrow-\infty} \frac{1}{f(x)}=. \qquad b. limx1f(x)=\lim _{x \rightarrow-1^{-}} f(x)= \qquad g. limx2+1f(x)=\lim _{x \rightarrow-2^{+}} \frac{1}{f(x)}= \qquad D] -\infty c. limx1+f(x)=\lim _{x \rightarrow 1^{+}} f(x)= \qquad d. limx+f(x)=\lim _{x \rightarrow+\infty} f(x)= \qquad h. limx1ef(x)=\lim _{x \rightarrow 1^{-}} e^{f(x)}= \qquad e. limx0[f(x)+3]=\lim _{x \rightarrow 0}[f(x)+3]= \qquad i. limx1+ef(x)=\lim _{x \rightarrow 1^{+}} e^{f(x)}= \qquad

Studdy Solution
h. Poiché f(x) f(x) diminuisce bruscamente quando x1 x \to 1^{-} , il limite di ef(x) e^{f(x)} è: limx1ef(x)=e=0 \lim_{x \to 1^{-}} e^{f(x)} = e^{-\infty} = 0
i. Poiché f(x) f(x) aumenta bruscamente quando x1+ x \to 1^{+} , il limite di ef(x) e^{f(x)} è: limx1+ef(x)=e+=+ \lim_{x \to 1^{+}} e^{f(x)} = e^{+\infty} = +\infty
Le uguaglianze completate sono: a. limxf(x)=1 \lim_{x \to -\infty} f(x) = 1 f. limx1f(x)=1 \lim_{x \to -\infty} \frac{1}{f(x)} = 1 b. limx1f(x)= \lim_{x \to -1^{-}} f(x) = -\infty c. limx1+f(x)=+ \lim_{x \to 1^{+}} f(x) = +\infty d. limx+f(x)=0 \lim_{x \to +\infty} f(x) = 0 h. limx1ef(x)=0 \lim_{x \to 1^{-}} e^{f(x)} = 0 i. limx1+ef(x)=+ \lim_{x \to 1^{+}} e^{f(x)} = +\infty

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