![f(x)=\displaystyle\frac{x^2+x-2}{x+1}=\frac{x(x+1)-2}{x+1}=x-\frac{2}{x+1} ](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdisplaystyle%5Cfrac%7Bx%5E2%2Bx-2%7D%7Bx%2B1%7D%3D%5Cfrac%7Bx%28x%2B1%29-2%7D%7Bx%2B1%7D%3Dx-%5Cfrac%7B2%7D%7Bx%2B1%7D%0A)
![\displaystyle\lim_{x\to\infty}\left\{f(x)-x \right\}=\lim_{x\to\infty}\frac{2}{x+1}=\lim_{x\to\infty}\frac{\displaystyle\frac{2}{x}}{1+\displaystyle\frac{1}{x}} =0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cleft%5C%7Bf%28x%29-x%20%5Cright%5C%7D%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B2%7D%7Bx%2B1%7D%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cdisplaystyle%5Cfrac%7B2%7D%7Bx%7D%7D%7B1%2B%5Cdisplaystyle%5Cfrac%7B1%7D%7Bx%7D%7D%0A%3D0)
Consequently, t<span>he limit of
![f(x)](https://tex.z-dn.net/?f=f%28x%29)
as x approaches infinity is
![x](https://tex.z-dn.net/?f=x)
.
In other words,
![f(x)](https://tex.z-dn.net/?f=f%28x%29)
approaches the line y=x,
</span><span>
so oblique asymptote is y=x.
I'm Japanese, if you find some mistakes in my English, please let me know.</span>
Answer:
150 units squared
Step-by-step explanation:
Answer:Just solve it. This is Algebra. Not that hard. Use Substation Method
Step-by-step explanation:
Answer:
a² + b² = c²
13² + 13² = c²
169 + 169 = c²
338 = c²
c = √338 or 18.385 or 13√2