Tính các giới hạn sau:
LG a
(underset{xrightarrow -3}{lim}) (frac{x^{2 }-1}{x+1});
Phương pháp giải:
Nếu hàm số (y=f(x)) xác định tại (x=x_0) thì (mathop {lim }limits_{x to {x_0}} fleft( x right) = fleft( {{x_0}} right)).
Nếu giới hạn hàm số có dạng vô định, tìm cách khử dạng vô định.
Lời giải chi tiết:
(underset{xrightarrow -3}{lim}) (dfrac{x^{2 }-1}{x+1}) ( = dfrac{{mathop {lim }limits_{x to – 3} left( {{x^2} – 1} right)}}{{mathop {lim }limits_{x to – 3} left( {x + 1} right)}} ) (= dfrac{{mathop {lim }limits_{x to – 3} {x^2} – mathop {lim }limits_{x to – 3} 1}}{{mathop {lim }limits_{x to – 3} x + mathop {lim }limits_{x to – 3} 1}}) = (dfrac{(-3)^{2}-1}{-3 +1} = -4).
LG b
(underset{xrightarrow -2}{lim}) (dfrac{4-x^{2}}{x + 2});
Lời giải chi tiết:
(underset{xrightarrow -2}{lim}) (dfrac{4-x^{2}}{x + 2}) = (underset{xrightarrow -2}{lim}) (dfrac{ (2-x)(2+x)}{x + 2}) = (underset{xrightarrow -2}{lim} (2-x) =2-(-2)= 4)
LG c
(underset{xrightarrow 6}{lim}) (dfrac{sqrt{x + 3}-3}{x-6})
Lời giải chi tiết:
(underset{xrightarrow 6}{lim}) (dfrac{sqrt{x + 3}-3}{x-6}) = (underset{xrightarrow 6}{lim}dfrac{(sqrt{x + 3}-3)(sqrt{x + 3}+3 )}{(x-6) (sqrt{x + 3}+3 )}) = (underset{xrightarrow 6}{lim}) (dfrac{x +3-9}{(x-6) (sqrt{x + 3}+3 )}) ( = mathop {lim }limits_{x to 6} dfrac{{x – 6}}{{left( {x – 6} right)left( {sqrt {x + 3} + 3} right)}} ) (= mathop {lim }limits_{x to 6} dfrac{1}{{sqrt {x + 3} + 3}} ) (= dfrac{1}{{mathop {lim }limits_{x to 6} left( {sqrt {x + 3} + 3} right)}} ) (= dfrac{1}{{mathop {lim }limits_{x to 6} left( {sqrt {x + 3} } right) + 3}} ) (= dfrac{1}{{sqrt {6 + 3} + 3}})= (dfrac{1}{6}).
LG d
(underset{xrightarrow +infty }{lim}) (dfrac{2x-6}{4-x})
Lời giải chi tiết:
(underset{xrightarrow +infty }{lim}) (dfrac{2x-6}{4-x}) ( = mathop {lim }limits_{x to + infty } dfrac{{xleft( {2 – dfrac{6}{x}} right)}}{{xleft( {dfrac{4}{x} – 1} right)}} ) (= mathop {lim }limits_{x to + infty } dfrac{{2 – dfrac{6}{x}}}{{dfrac{4}{x} – 1}} ) (= dfrac{{2 – mathop {lim }limits_{x to + infty } dfrac{6}{x}}}{{mathop {lim }limits_{x to + infty } dfrac{4}{x} – 1}} ) (= dfrac{{2 – 0}}{{0 – 1}}) ( = -2)
LG e
(underset{xrightarrow +infty }{lim}) (dfrac{17}{x^{2}+1})
Lời giải chi tiết:
(underset{xrightarrow +infty }{lim}) (dfrac{17}{x^{2}+1} = 0) vì:
(underset{xrightarrow +infty }{lim}) ((x^2+ 1) =) (underset{xrightarrow +infty }{lim} x^2( 1 + dfrac{1}{x^{2}}) = +∞)
Cách khác:
(mathop {lim }limits_{x to + infty } dfrac{{17}}{{{x^2} + 1}}) ( = mathop {lim }limits_{x to + infty } dfrac{{{x^2}.dfrac{{17}}{{{x^2}}}}}{{{x^2}.left( {1 + dfrac{1}{{{x^2}}}} right)}} ) (= mathop {lim }limits_{x to + infty } dfrac{{dfrac{{17}}{{{x^2}}}}}{{1 + dfrac{1}{{{x^2}}}}} ) (= dfrac{{mathop {lim }limits_{x to + infty } dfrac{{17}}{{{x^2}}}}}{{1 + mathop {lim }limits_{x to + infty } dfrac{1}{{{x^2}}}}} ) (= dfrac{0}{{1 + 0}} = 0)
LG f
(underset{xrightarrow +infty }{lim}) (dfrac{-2x^{2}+x -1}{3 +x})
Lời giải chi tiết:
(underset{xrightarrow +infty }{lim}) (dfrac{-2x^{2}+x -1}{3 +x}) (= mathop {lim }limits_{x to + infty } dfrac{{{x^2}left( { – 2 + dfrac{1}{x} – dfrac{1}{{{x^2}}}} right)}}{{{x^2}left( {dfrac{3}{{{x^2}}} + dfrac{1}{x}} right)}}) (=underset{xrightarrow +infty }{lim}dfrac{-2+dfrac{1}{x} -dfrac{1}{x^{2}}}{dfrac{3}{x^{2}} +dfrac{1}{x}} )
Vì (mathop {lim }limits_{x to + infty } left( {dfrac{3}{{{x^2}}} + dfrac{1}{x}} right) = 0); ({dfrac{3}{{{x^2}}} + dfrac{1}{x}}>0) khi (x to + infty)
và (mathop {lim }limits_{x to + infty } left( { – 2 + dfrac{1}{x} – dfrac{1}{{{x^2}}}} right) ) (= – 2 + mathop {lim }limits_{x to + infty } dfrac{1}{x} – mathop {lim }limits_{x to + infty } dfrac{1}{{{x^2}}}) ( = – 2 + 0 – 0 = – 2 < 0)
Vậy (underset{xrightarrow +infty }{lim}) (dfrac{-2x^{2}+x -1}{3 +x})(=underset{xrightarrow +infty }{lim}dfrac{-2+dfrac{1}{x} -dfrac{1}{x^{2}}}{dfrac{3}{x^{2}} +dfrac{1}{x}} ) (=-infty )
Cách khác:
(mathop {lim }limits_{x to + infty } dfrac{{ – 2{x^2} + x – 1}}{{3 + x}} ) (= mathop {lim }limits_{x to + infty } dfrac{{{x^2}left( { – 2 + dfrac{1}{x} – dfrac{1}{{{x^2}}}} right)}}{{xleft( {dfrac{3}{x} + 1} right)}}) ( = mathop {lim }limits_{x to + infty } left[ {x.dfrac{{ – 2 + dfrac{1}{x} – dfrac{1}{{{x^2}}}}}{{dfrac{3}{x} + 1}}} right])
Mà (mathop {lim }limits_{x to + infty } x = + infty )
và (mathop {lim }limits_{x to + infty } dfrac{{ – 2 + dfrac{1}{x} – dfrac{1}{{{x^2}}}}}{{dfrac{3}{x} + 1}} ) (= dfrac{{ – 2 + mathop {lim }limits_{x to + infty } dfrac{1}{x} – mathop {lim }limits_{x to + infty } dfrac{1}{{{x^2}}}}}{{mathop {lim }limits_{x to + infty } dfrac{3}{x} + 1}} ) (= dfrac{{ – 2 + 0 – 0}}{{0 + 1}} = – 2 < 0)
Nên (underset{xrightarrow +infty }{lim}) (dfrac{-2x^{2}+x -1}{3 +x})(=-infty )