Kết quả của giới hạn \(\mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{x - 15}}{{x - 2}}\) là:

\(\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) = L\)

\(\mathop {\lim }\limits_{x \to  - \infty } f\left( x \right) = L\)

\(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L\)

\(\mathop {\lim }\limits_{x \to L} f\left( x \right) = {x_0}\)

\(\dfrac{1}{5}.\)

\(\sqrt 5 .\)

\(\dfrac{1}{{\sqrt 5 }}.\)

\(5.\)

\(\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = L\)

\(\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = M\)

\(\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = L - M\)                           

\(\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = M + L\)

\(0.\)

\(1.\)

\(2.\)

\(3.\)

\(\mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = L\)           

\(\mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right) = L\)

\(\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) = L\)

\(\mathop {\lim }\limits_{x \to  - \infty } f\left( x \right) = L\)

\(1.\)

\( - \infty .\)

\(0.\)

\( + \infty .\)

\(\mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = L\)

\(\mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) =  - L\)        

\(\mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right) =  - L\)

$\mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = - \mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right)$