Advertisements (Quảng cáo)
Tìm các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to 3} \left( {{1 \over x} – {1 \over 3}} \right){1 \over {{{\left( {x – 3} \right)}^3}}}\) b) \(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} {{4{x^4} – 3} \over {2{x^2} + 3x – 2}}\)
a) Với mọi \(x \ne 3,\)
\(\left( {{1 \over x} – {1 \over 3}} \right){1 \over {{{\left( {x – 3} \right)}^3}}} = {{3 – x} \over {3x}}.{1 \over {{{\left( {x – 3} \right)}^3}}} = \left( { – {1 \over {3x}}} \right).{1 \over {{{\left( {x – 3} \right)}^2}}}.\)
Vì \(\mathop {\lim }\limits_{x \to 3} \left( { – {1 \over {3x}}} \right) = – {1 \over 9} < 0\) và \(\mathop {\lim }\limits_{x \to 3} {1 \over {{{\left( {x – 3} \right)}^2}}} = + \infty \) nên
\(\mathop {\lim }\limits_{x \to 3} \left( {{1 \over x} – {1 \over 3}} \right){1 \over {{{\left( {x – 3} \right)}^3}}} = – \infty ;\)
b) \({{4{x^4} – 3} \over {2{x^2} + 3x – 2}} = {{4{x^4} – 3} \over {2x – 1}}.{1 \over {x + 2}}\)
Vì \(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} {{4{x^4} – 3} \over {2x – 1}} = {{ – 61} \over 5} < 0\) và \(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} {1 \over {x + 2}} = + \infty \) nên
Advertisements (Quảng cáo)
\(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} {{4{x^4} – 3} \over {2{x^2} + 3x – 2}} = – \infty .\)
Cách giải khác
Vì \(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} \left( {4{x^4} – 3} \right) = 61 > 0,\)
\(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} \left( {2{x^2} + 3x – 2} \right) = 0\) và \(2{x^2} + 3x – 2 < 0\)
Với \( – 2 < x < {1 \over 2}\) nên
\(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} {{4{x^4} – 3} \over {2{x^2} + 3x – 2}} = – \infty .\)