Granice

[Invisibl3]

Witam, bardzo proszę o pomoc w rozwiązaniu tych zadań

\(1. \Lim _{x\to \infty }\left(\sqrt{5^4+x^2-1}-\sqrt{5x^4+7x+3}\right)\\

2. \Lim _{x\to 2}\frac{\sqrt{2x^2+1}-\sqrt{x+7}}{x^2-4}\\

3.\Lim _{x\to 0}\left(2^5-\frac{1}{^{^{x^2}}}\right)\\

4.\Lim _{x\to 1}\left(\frac{4x+3}{3x+4}\right)^{\frac{1}{1-x}}\)


pozdrawiam serdecznie

[eresh]

1. \(\lim _{x\to \infty }\left(\sqrt{5x^4+x^2-1}-\sqrt{5x^4+7x+3}\right)
\)

\(\Lim _{x\to \infty }\left(\sqrt{5x^4+x^2-1}-\sqrt{5x^4+7x+3}\right)=\Lim_{x\to \infty}\frac{5x^4+x^2-1-5x^4-7x-3}{\sqrt{5x^4+x^2-1}+\sqrt{5x^4+7x+3}}=\Lim_{x\to \infty}\frac{x^2-7x-4}{\sqrt{5x^4+x^2-1}+\sqrt{5x^4+7x+3}}=\\=\Lim_{x\to\infty}\frac{1-\frac{7}{x}-\frac{4}{x^2}}{\sqrt{5+\frac{1}{x^2}-\frac{1}{x^4}}+\sqrt{5+\frac{7}{x^3}+\frac{3}{x^4}}}=\frac{1}{2\sqrt{5}}\)

[eresh]

2. \(\lim _{x\to 2}\frac{\sqrt{2x^2+1}-\sqrt{x+7}}{x^2-4}\)

\(\Lim_{x\to 2}\frac{\sqrt{2x^2+1}-\sqrt{x+7}}{x^2-4}=\Lim_{x\to 2}\frac{(\sqrt{2x^2+1}-\sqrt{x+7})(\sqrt{2x^2+1}+\sqrt{x+7})}{(x-2)(x+2)\sqrt{2x^2+1}+\sqrt{x+7})}=\Lim_{x\to 2}\frac{2x^2-x-6}{(x-2)(x+2)(\sqrt{2x^2+1}+\sqrt{x+7})}=\\=\Lim_{x\to 2}\frac{2(x-2)(2x+3)}{(x-2)(x+2)(\sqrt{2x^2+1}+\sqrt{x+7})}=\Lim_{x\to 2}\frac{(2x+3)}{(x+2)(\sqrt{2x^2+1}+\sqrt{x+7})}=\frac{7}{24}\)

[eresh]

3.\lim _{x\to 0}\left(2^5-\frac{1}{^{^{x^2}}}\right)

\(\Lim _{x\to 0}\left(2^5-\frac{1}{^{^{x^2}}}\right)\)
tak to ma wyglądać?

[eresh]

4.\(\lim _{x\to 1}\left(\frac{4x+3}{3x+4}\right)^{\frac{1}{1-x}}
\)

\(\Lim _{x\to 1}\left(\frac{4x+3}{3x+4}\right)^{\frac{1}{1-x}}=\Lim_{x\to 1}e^{\frac{\ln\frac{4x+3}{3x+4}}{1-x}}=e^{\Lim_{x\to 1}\frac{\ln\frac{4x+3}{3x+4}}{1-x}}=_H e^{\Lim_{x\to 1}\frac{\frac{7}{(4x+3)(3x+4}}{-1}}=e^{-\frac{1}{7}}\)

[Invisibl3]

Rewelacja, dziekuję za pomoc