Cho hàm số \(f\left( x \right) = {{2{x^2} – 15x + 12} \over {{x^2} – 5x + 4}}\) có đồ thị như hình 4
a) Dựa vào đồ thị, dự đoán giới hạn của hàm \(f\left( x \right)\) số khi \(x \to {1^ + }{\rm{ }};{\rm{ }}x \to {1^ – }{\rm{ }};{\rm{ }}x \to {4^ + }{\rm{ }};{\rm{ }}x \to {4^ – }{\rm{ }};{\rm{ }}x \to + \infty {\rm{ }};{\rm{ }}x \to – \infty \)
b) Chứng minh dự đoán trên.
a) Dự đoán :
\(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = + \infty {\rm{ ; }}\mathop {\lim }\limits_{x \to {1^ – }} f\left( x \right) = – \infty {\rm{ ; }}\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right) = – \infty {\rm{ ;}} \cr
& {\rm{ }}\mathop {\lim }\limits_{x \to {4^ – }} f\left( x \right) = + \infty {\rm{ ;}}\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = 2{\rm{ ; }}\mathop {\lim }\limits_{x \to – \infty } f\left( x \right) = 2. \cr} \)
b) Ta có
\(\mathop {\lim }\limits_{x \to {1^ + }} \left( {2{x^2} – 15x + 12} \right) = – 1 < 0,{\rm{ }}\mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} – 5x + 4} \right) = 0\)
và \({x^2} – 5x + 4 < 0\) với mọi \(x \in \left( {1;4} \right)\) nên \(\mathop {\lim }\limits_{x \to {1^ + }} {{2{x^2} – 15x + 12} \over {{x^2} – 5x + 4}} = + \infty \)
Vì
\(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ – }} \left( {2{x^2} – 15x + 12} \right) = – 1 < 0, \cr
& \mathop {\lim }\limits_{x \to {1^ – }} \left( {{x^2} – 5x + 4} \right) = 0 \cr} \)
Advertisements (Quảng cáo)
và \({x^2} – 5x + 4 > 0\) với mọi x < 1 nên \(\mathop {\lim }\limits_{x \to {1^ – }} {{2{x^2} – 15x + 12} \over {{x^2} – 5x + 4}} = – \infty \)
Vì
\(\eqalign{
& \mathop {\lim }\limits_{x \to {4^ + }} \left( {2{x^2} – 15x + 12} \right) = – 16 < 0, \cr
& \mathop {\lim }\limits_{x \to {4^ + }} \left( {{x^2} – 5x + 4} \right) = 0 \cr} \)
và \({x^2} – 5x + 4 > 0\) với mọi x > 4 nên \(\mathop {\lim }\limits_{x \to {4^ + }} {{2{x^2} – 15x + 12} \over {{x^2} – 5x + 4}} = – \infty \)
Vì
\(\eqalign{
& \mathop {\lim }\limits_{x \to {4^ – }} \left( {2{x^2} – 15x + 12} \right) = – 16 < 0, \cr
& \mathop {\lim }\limits_{x \to {4^ – }} \left( {{x^2} – 5x + 4} \right) = 0 \cr} \)
và \({x^2} – 5x + 4 < 0\) với mọi \(x \in \left( {1;4} \right)\) nên \(\mathop {\lim }\limits_{x \to {4^ – }} {{2{x^2} – 15x + 12} \over {{x^2} – 5x + 4}} = + \infty\) ;
\(\mathop {\lim }\limits_{x \to + \infty } {{2{x^2} – 15x + 12} \over {{x^2} – 5x + 4}} = \mathop {\lim }\limits_{x \to + \infty } {{2 – {{15} \over x} + {{12} \over {{x^2}}}} \over {1 – {5 \over x} + {4 \over {{x^2}}}}} = 2\)
\(\mathop {\lim }\limits_{x \to – \infty } {{2{x^2} – 15x + 12} \over {{x^2} – 5x + 4}} = \mathop {\lim }\limits_{x \to – \infty } {{2 – {{15} \over x} + {{12} \over {{x^2}}}} \over {1 – {5 \over x} + {4 \over {{x^2}}}}} = 2\)