Tìm giới hạn của các hàm số sau
a) \(f(x) = \frac{{{x^2} - 2x - 3}}{{x - 1}}\) khi \(x\to 3\)
b) \(h(x) = \frac{{2{x^3} + 15}}{{{{(x + 2)}^2}}}\) khi \(x\to -2\)
c) \(k(x) = \sqrt {4{x^2} - x + 1} \) khi \(x \to - \infty \)
d) \(h(x) = \frac{{x - 15}}{{x + 2}}\) khi \(x \to - {2^ + }\) và
a) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 2x - 3}}{{x - 1}} = \frac{{{3^2} - 2.3 - 3}}{{3 - 1}} = 0\)
b) \(\mathop {\lim }\limits_{x \to - 2} h\left( x \right) = \mathop {\lim }\limits_{x \to - 2} \frac{{2{x^3} + 15}}{{{{\left( {x + 2} \right)}^2}}} = - \infty \)
Vì \(\mathop {\lim }\limits_{x \to - 2} \left( {2{x^3} + 15} \right) = - 1 < 0\) và \(\mathop {\lim }\limits_{x \to - 3} {\left( {x + 2} \right)^2} = 0\)
c)
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } k\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \sqrt {4{x^2} - x + 1} \\
= \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2}\left( {4 - \frac{1}{x} + \frac{1}{{{x^2}}}} \right)} \\
= \mathop {\lim }\limits_{x \to - \infty } \left| x \right|\sqrt {4 - \frac{1}{x} + \frac{1}{{{x^2}}}} = + \infty
\end{array}\)
d) \(\mathop {\lim }\limits_{x \to - {2^ + }} h\left( x \right) = \mathop {\lim }\limits_{x \to - {2^ + }} \frac{{x - 15}}{{x + 2}} = - \infty \)
Vì \(\mathop {\lim }\limits_{x \to - {2^ + }} (x - 15) = - 17 < 0\) và \(\mathop {\lim }\limits_{x \to - {2^ + }} \left( {x + 2} \right) = 0,x + 2 > 0,\forall x > - 2\)
\(\mathop {\lim }\limits_{x \to - {2^ + }} \left( {x + 2} \right) = 0,\mathop {\lim }\limits_{x \to - {2^ - }} h\left( x \right) = \mathop {\lim }\limits_{x \to - {2^ - }} \frac{{x - 15}}{{x + 2}} = + \infty ,x + 2 > 0,\forall x > - 2\)
Vì \(\mathop {\lim }\limits_{x \to - {2^ - }} (x - 15) = - 17 < 0\) và \(\mathop {\lim }\limits_{x \to - {2^ - }} \left( {x + 2} \right) = 0,x + 2 < 0,\forall x < - 2\)
-- Mod Toán 11