(11) Granic ciąg dalszy

[melon]

Oblicz granice nie korzystając z reguły de l'Hospitala

\(\lim_{x\to 1} \frac{|tg(x-1)|}{(x-1)^2}\)

\(\lim_{x\to \frac{1}{2} } \frac{\arcsin(1-2x)}{4x^2-1}\)

\(\lim_{x\to 0} \sqrt[x]{1+\sin x}\)

\(\lim_{x\to 0} (1+kx)^{ \frac{n}{x}}\)

\(\lim_{x\to 0 } \frac{arctg x}{x}\)

[octahedron]

\(\lim_{x\to 1}\frac{|tg(x-1)|}{(x-1)^2}=\lim_{x\to 1}\frac{|tg(x-1)|}{|x-1|}\cdot\frac{1}{|x-1|}=1\cdot \frac{1}{0^+}=\infty\)

[octahedron]

\(\lim_{x\to \frac{1}{2} } \frac{\arcsin(1-2x)}{4x^2-1}=\lim_{x\to \frac{1}{2} } -\frac{\arcsin(1-2x)}{1-2x}\cdot\frac{1}{2x+1}=\lim_{z\to 0} -\frac{z}{\sin z}\cdot\lim_{x\to \frac{1}{2} }\frac{1}{2x+1}=-1\cdot\frac{1}{2}=-\frac{1}{2}\)

[octahedron]

\(\lim_{x\to 0} \sqrt[x]{1+\sin x}=\lim_{x\to 0}\(1+\sin x\)^{\frac{1}{x}}=\lim_{x\to 0}\(1+\sin x\)^{\frac{1}{\sin x}\cdot \frac{\sin x}{x}}=e^1=e\)

[octahedron]

\(\lim_{x\to 0} (1+kx)^{ \frac{n}{x}}=\lim_{x\to 0} (1+kx)^{ \frac{1}{kx}\cdot\frac{k}{n}}=e^{\frac{k}{n}}\)

[octahedron]

\(arctg x=z\Rightarrow x=tg z
\lim_{x\to 0 } \frac{arctg x}{x}=\lim_{z\to 0 } \frac{z}{tg z}=1\)