Obliczyć granice funkcji

[magdabp]

\(\displaystyle{ (a) \lim_{x \to +\infty} \frac{\sqrt{x^2+1}}{x+1}}\)

\(\displaystyle{ (b) \lim_{x \to +\infty} (\frac{3x+1}{3x-2})^{2x}}\)

\(\displaystyle{ (c) \lim_{x \to 0} \frac{tgx-sinx}{sin^3x}}\)

\(\displaystyle{ (d) \lim_{x \to 1} \frac{x^2-\sqrt{x}}{1-\sqrt{x}}}\)

\(\displaystyle{ (e) \lim_{x \to +\infty} (\frac{x^2-1}{x^2})^x}\)

\(\displaystyle{ (f) \lim_{x \to 1} \frac{x^3-1}{x^4+3x^2-1}}\)

\(\displaystyle{ (g) \lim_{x \to 0} \frac{4x^2-3x}{2x^2-9x}}\)

[dem]

2.
\(\displaystyle{ \lim_{x \to +\infty} (\frac{3x+1}{3x-2})^{2x}=
\lim_{x \to +\infty} (\frac{3x-2+3}{3x-2})^{2x}=\lim_{x \to +\infty} (1+\frac{1}{\frac{3x-2}{3}})^{\frac{6x}{3x-2}}=e^2}\)

[andkom]

d)
\(\displaystyle{ \lim_{x\to1}\frac{x^2-\sqrt x}{1-\sqrt x}=
\lim_{x\to1}\frac{-\sqrt x(1-(\sqrt x)^3)}{1-\sqrt x}=\\
=\lim_{x\to1}\frac{-\sqrt x(1-\sqrt x)(1+\sqrt x+(\sqrt x)^2)}{1-\sqrt x}=\\
=\lim_{x\to1}-\sqrt x(1+\sqrt x+(\sqrt x)^2)=-\sqrt 1(1+\sqrt 1+(\sqrt 1)^2)=-3}\)

[ Dodano: 12 Października 2007, 22:31 ]
g)
\(\displaystyle{ \lim_{x\to0}\frac{4x^2-3x}{2x^2-9x}
=\lim_{x\to0}\frac{x(4x-3)}{x(2x-9)}=\\
=\lim_{x\to0}\frac{4x-3}{2x-9}
=\frac{4\cdot0-3}{2\cdot0-9}=\frac13}\)