Question

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  - {1^ + }} \frac{1}{{x + 1}}\);              

b) \(\mathop {\lim }\limits_{x \to  - \infty } \left( {1 - {x^2}} \right)\);

c) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{x}{{3 - x}}\).

Answer 1

a: \(\lim\limits_{x\rightarrow-1^+}x+1=0\)

=>\(\lim\limits_{x\rightarrow-1^+}\dfrac{1}{x+1}=+\infty\)

b: \(\lim\limits_{x\rightarrow-\infty}1-x^2=\lim\limits_{x\rightarrow-\infty}\left[x^2\left(\dfrac{1}{x^2}-1\right)\right]\)

\(=-\infty\)

c: \(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=\lim\limits_{x\rightarrow3^-}=\dfrac{-x}{x-3}\)

\(\lim\limits_{x\rightarrow3^-}x-3=0\)

\(\lim\limits_{x\rightarrow3^-}-x=3>0\)

=>\(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=+\infty\)