Во первых очевидно, что
![\lim\limits_{x\to \infty}\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}} =1\\\lim\limits_{x\to \infty}\frac{x+1}{x-1} =1](https://tex.z-dn.net/?f=%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Cfrac%7Bx%2B%5Csqrt%7Bx%5E2%2B1%7D%7D%7Bx%2B%5Csqrt%7Bx%5E2-1%7D%7D%20%3D1%5C%5C%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Cfrac%7Bx%2B1%7D%7Bx-1%7D%20%3D1)
Поэтому, при ![x \to \infty](https://tex.z-dn.net/?f=x%20%5Cto%20%5Cinfty)
![\ln(\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}})=\ln(1+(\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}-1)) \sim \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}-1=\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} \\\ln^2\frac{x+1}{x-1} =\ln^2(1+(\frac{x+1}{x-1}-1)) \sim (\frac{x+1}{x-1}-1)^2=\frac{4}{(x-1)^2}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7Bx%2B%5Csqrt%7Bx%5E2%2B1%7D%7D%7Bx%2B%5Csqrt%7Bx%5E2-1%7D%7D%29%3D%5Cln%281%2B%28%5Cfrac%7Bx%2B%5Csqrt%7Bx%5E2%2B1%7D%7D%7Bx%2B%5Csqrt%7Bx%5E2-1%7D%7D-1%29%29%20%5Csim%20%5Cfrac%7Bx%2B%5Csqrt%7Bx%5E2%2B1%7D%7D%7Bx%2B%5Csqrt%7Bx%5E2-1%7D%7D-1%3D%5Cfrac%7B%5Csqrt%7Bx%5E2%2B1%7D-%5Csqrt%7Bx%5E2-1%7D%7D%7Bx%2B%5Csqrt%7Bx%5E2-1%7D%7D%20%5C%5C%5Cln%5E2%5Cfrac%7Bx%2B1%7D%7Bx-1%7D%20%3D%5Cln%5E2%281%2B%28%5Cfrac%7Bx%2B1%7D%7Bx-1%7D-1%29%29%20%5Csim%20%28%5Cfrac%7Bx%2B1%7D%7Bx-1%7D-1%29%5E2%3D%5Cfrac%7B4%7D%7B%28x-1%29%5E2%7D)
Перепишем исходный предел, использовав эти эквивалентности:
![=\lim\limits_{x\to \infty}\frac{(\sqrt{x^2+1}-\sqrt{x^2-1})(x-1)^2}{4(x+\sqrt{x^2+1})} =\frac{1}{4} \lim\limits_{x\to \infty}\frac{x^2(\sqrt{x^2+1}-\sqrt{x^2-1})}{x+\sqrt{x^2+1}}=\\ =\frac{1}{2} \lim\limits_{x\to \infty}\frac{x^2}{(x+\sqrt{x^2+1})(\sqrt{x^2+1}+\sqrt{x^2-1})} =\frac{1}{2} \lim\limits_{x\to \infty}\frac{x^2}{x\sqrt{x^2+1}+x^2+1+x\sqrt{x^2-1}+\sqrt{x^4-1}} =\\](https://tex.z-dn.net/?f=%3D%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Cfrac%7B%28%5Csqrt%7Bx%5E2%2B1%7D-%5Csqrt%7Bx%5E2-1%7D%29%28x-1%29%5E2%7D%7B4%28x%2B%5Csqrt%7Bx%5E2%2B1%7D%29%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Cfrac%7Bx%5E2%28%5Csqrt%7Bx%5E2%2B1%7D-%5Csqrt%7Bx%5E2-1%7D%29%7D%7Bx%2B%5Csqrt%7Bx%5E2%2B1%7D%7D%3D%5C%5C%20%3D%5Cfrac%7B1%7D%7B2%7D%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Cfrac%7Bx%5E2%7D%7B%28x%2B%5Csqrt%7Bx%5E2%2B1%7D%29%28%5Csqrt%7Bx%5E2%2B1%7D%2B%5Csqrt%7Bx%5E2-1%7D%29%7D%20%3D%5Cfrac%7B1%7D%7B2%7D%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Cfrac%7Bx%5E2%7D%7Bx%5Csqrt%7Bx%5E2%2B1%7D%2Bx%5E2%2B1%2Bx%5Csqrt%7Bx%5E2-1%7D%2B%5Csqrt%7Bx%5E4-1%7D%7D%20%3D%5C%5C)
![=\frac{1}{2} \lim\limits_{x\to \infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}+1+\frac{1}{x^2}+\sqrt{1-\frac{1}{x^2}}+\sqrt{1-\frac{1}{x^4}}}=\frac{1}{2} *\frac{1}{4} =\frac{1}{8}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Cfrac%7B1%7D%7B%5Csqrt%7B1%2B%5Cfrac%7B1%7D%7Bx%5E2%7D%7D%2B1%2B%5Cfrac%7B1%7D%7Bx%5E2%7D%2B%5Csqrt%7B1-%5Cfrac%7B1%7D%7Bx%5E2%7D%7D%2B%5Csqrt%7B1-%5Cfrac%7B1%7D%7Bx%5E4%7D%7D%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2A%5Cfrac%7B1%7D%7B4%7D%20%3D%5Cfrac%7B1%7D%7B8%7D)
1)√5>2, /√5-2/=√5-2
2)√4=2, /√4-2/=√4-2=2-2=0
3)√3<2,/√3-2/=2-√3
(x+5)(x-k)<0
Рассмотрим 2 случая, k>-5 и k<-5
1) k>-5 + - +
____________(-5)/////////////////////////////////////////// (k)__________
-4 -3 -2 -1 k=0
2) k<-5 + - +
____________(k)//////////////////////////////////////////// (-5) __________
k=-10 -9 -8 -7 -6
Ответ: при k=-10 и k=0