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a) Với x > 0 bất kì và \(h = x - {x_0}\) ta có
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{x_0} + h} \right) - \ln {x_0}}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}.{x_0}}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x_0}}}.\mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}}} = \frac{1}{{{x_0}}}\end{array}\)
Vậy hàm số \(y = \ln x\) có đạo hàm là hàm số \(y' = \frac{1}{x}\)
b) Ta có \({\log _a}x = \frac{{\ln x}}{{\ln a}}\) nên \(\left( {{{\log }_a}x} \right)' = \left( {\frac{{\ln x}}{{\ln a}}} \right)' = \frac{1}{{x\ln a}}\)
a) Với bất kì \({x_0} \in \mathbb{R}\), ta có:
\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{e^x} - {e^{{x_0}}}}}{{x - {x_0}}}\)
Đặt \(x = {x_0} + \Delta x\). Ta có:
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0} + \Delta x}} - {e^{{x_0}}}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0}}}.{e^{\Delta x}} - {e^{{x_0}}}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0}}}.\left( {{e^{\Delta x}} - 1} \right)}}{{\Delta x}}\\ & = \mathop {\lim }\limits_{\Delta x \to 0} {e^{{x_0}}}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{\Delta x}} - 1}}{{\Delta x}} = {e^{{x_0}}}.1 = {e^{{x_0}}}\end{array}\)
Vậy \({\left( {{e^x}} \right)^\prime } = {e^x}\) trên \(\mathbb{R}\).
b) Với bất kì \({x_0} > 0\), ta có:
\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln {\rm{x}} - \ln {{\rm{x}}_0}}}{{x - {x_0}}}\)
Đặt \(x = {x_0} + \Delta x\). Ta có:
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{x_0} + \Delta x} \right) - \ln {{\rm{x}}_0}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {\frac{{{x_0} + \Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\Delta x}}\\ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}}.\frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}}\end{array}\)
Đặt \(\frac{{\Delta x}}{{{x_0}}} = t\). Lại có: \(\mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}} = \frac{1}{{{x_0}}};\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\ln \left( {1 + t} \right)}}{t} = 1\)
Vậy \(f'\left( {{x_0}} \right) = \frac{1}{{{x_0}}}.1 = \frac{1}{{{x_0}}}\)
Vậy \({\left( {\ln x} \right)^\prime } = \frac{1}{x}\) trên khoảng \(\left( {0; + \infty } \right)\).
a: \(\lim\limits_{x\rightarrow-1^+}x+1=0\)
=>\(\lim\limits_{x\rightarrow-1^+}\dfrac{1}{x+1}=+\infty\)
b: \(\lim\limits_{x\rightarrow-\infty}1-x^2=\lim\limits_{x\rightarrow-\infty}\left[x^2\left(\dfrac{1}{x^2}-1\right)\right]\)
\(=-\infty\)
c: \(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=\lim\limits_{x\rightarrow3^-}=\dfrac{-x}{x-3}\)
\(\lim\limits_{x\rightarrow3^-}x-3=0\)
\(\lim\limits_{x\rightarrow3^-}-x=3>0\)
=>\(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=+\infty\)
a) \(\sin \left( {x + h} \right) - \sin x = 2\cos \frac{{2x + h}}{2}.\sin \frac{h}{2}\)
b) Với \({x_0}\) bất kì, ta có:
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin x - \sin {x_0}}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{2\cos \frac{{x + {x_0}}}{2}.\sin \frac{{x - {x_0}}}{2}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin \frac{{x - {x_0}}}{2}}}{{\frac{{x - {x_0}}}{2}}}.\mathop {\lim }\limits_{x \to {x_0}} \cos \frac{{x + {x_0}}}{2} = \cos {x_0}\end{array}\)
Vậy hàm số y = sin x có đạo hàm là hàm số \(y' = \cos x\)
a) \(\mathop {\lim }\limits_{x \to - 1} \left( {3{x^2} - x + 2} \right) = \mathop {\lim }\limits_{x \to - 1} \left( {3{x^2}} \right) - \mathop {\lim }\limits_{x \to - 1} x + \mathop {\lim }\limits_{x \to - 1} 2\)
\( = 3\mathop {\lim }\limits_{x \to - 1} \left( {{x^2}} \right) - \mathop {\lim }\limits_{x \to - 1} x + \mathop {\lim }\limits_{x \to - 1} 2 = 3.{\left( { - 1} \right)^2} - \left( { - 1} \right) + 2 = 6\)
b) \(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \left( {x + 4} \right) = \mathop {\lim }\limits_{x \to 4} x + \mathop {\lim }\limits_{x \to 4} 4 = 4 + 4 = 8\)
c) \(\mathop {\lim }\limits_{x \to 2} \frac{{3 - \sqrt {x + 7} }}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3 - \sqrt {x + 7} } \right)\left( {3 + \sqrt {x + 7} } \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{3^2} - \left( {x + 7} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}}\)
\( = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{3 + \sqrt {x + 7} }}\)
\( = \frac{{\mathop {\lim }\limits_{x \to 2} \left( { - 1} \right)}}{{\mathop {\lim }\limits_{x \to 2} 3 + \sqrt {\mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 7} }} = \frac{{ - 1}}{{3 + \sqrt {2 + 7} }} = - \frac{1}{6}\)
a: \(\lim\limits_{x\rightarrow-2}x^2-7x+4=\left(-2\right)^2-7\cdot\left(-2\right)+4=22\)
b: \(\lim\limits_{x\rightarrow3}\dfrac{x-3}{x^2-9}=\lim\limits_{x\rightarrow3}\dfrac{1}{x+3}=\dfrac{1}{3+3}=\dfrac{1}{6}\)
c: \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-x-8}{3+\sqrt{x+8}}\cdot\dfrac{1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{-1}{3+\sqrt{x+8}}\)
\(=-\dfrac{1}{6}\)
a) \(\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} + 5x - 2} \right) = \mathop {\lim }\limits_{x \to - 2} {x^2} + \mathop {\lim }\limits_{x \to - 2} \left( {5x} \right) - \mathop {\lim }\limits_{x \to - 2} 2\)
\( = \mathop {\lim }\limits_{x \to - 2} {x^2} + 5\mathop {\lim }\limits_{x \to - 2} x - \mathop {\lim }\limits_{x \to - 2} 2 = {\left( { - 2} \right)^2} + 5.\left( { - 2} \right) - 2 = - 8\)
b) \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right) = \mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1 = 1 + 1 = 2\)
a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} - 4.2 + 3 = - 1\)
b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 1\)
c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x + 1}} = \frac{1}{{\sqrt 1 + 1}} = \frac{1}{2}\)
a) Đặt \(f\left( x \right) = 2{x^2} - x\).
Hàm số \(y = f\left( x \right)\) xác định trên \(\mathbb{R}\).
Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì thỏa mãn \({x_n} \to 3\) khi \(n \to + \infty \). Ta có:
\(\lim f\left( {{x_n}} \right) = \lim \left( {2x_n^2 - {x_n}} \right) = 2.\lim x_n^2 - \lim {x_n} = {2.3^2} - 3 = 15\).
Vậy \(\mathop {\lim }\limits_{x \to 3} \left( {2{x^2} - x} \right) = 15\).
b) Đặt \(f\left( x \right) = \frac{{{x^2} + 2x + 1}}{{x + 1}}\).
Hàm số \(y = f\left( x \right)\) xác định trên \(\mathbb{R}\).
Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì thỏa mãn \({x_n} \to - 1\) khi \(n \to + \infty \). Ta có:
\(\lim f\left( {{x_n}} \right) = \lim \frac{{x_n^2 + 2{x_n} + 1}}{{{x_n} + 1}} = \lim \frac{{{{\left( {{x_n} + 1} \right)}^2}}}{{{x_n} + 1}} = \lim \left( {{x_n} + 1} \right) = \lim {x_n} + 1 = - 1 + 1 = 0\).
Vậy \(\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 2x + 1}}{{x + 1}} = 0\).
a) Ta có \(t = \frac{1}{x},\) nên khi x tiến đến 0 thì t tiến đến dương vô cùng do đó
\(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = \mathop {\lim }\limits_{t \to + \infty } {\left( {1 + \frac{1}{t}} \right)^t} = e\)
b) \(\ln y = \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \frac{1}{x}\ln \left( {1 + x} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\)
c) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)