Calculus: Limits

$$x$$  is approaching $$a$$  

$$x$$  equals to $$a$$  

mapping from $$x$$  to $$a$$  

substitute $$x$$  by $$a$$  

$$f\left(a\right)=L$$  

$$f\left(x\right)\rightarrow L$$  

$$f\left(x\right)\rightarrow L$$  when $$x\rightarrow a$$  

$$f\left(a\right)\rightarrow L$$  

$$\lim_{x\rightarrow a}f\left(x\right)$$  is a finite number

$$\lim_{x\rightarrow a}f\left(x\right)=f\left(a\right)$$  

$$\lim_{x\rightarrow a}f\left(x\right)>0$$  

$$\lim_{x\rightarrow a}f\left(x\right)\ne\infty$$  

$$\lim_{x\rightarrow-1}\frac{x+1}{x^2+1}$$  

$$\lim_{x\rightarrow-1}\frac{x+1}{x^2-1}$$  

$$\lim_{x\rightarrow\infty}\frac{x+1}{x^2-1}$$  

$$\lim_{x\rightarrow1}\frac{x-1}{x^2-1}$$  

$$\sqrt{x+1}+1$$

$$\sqrt{x+1}-1$$

$$\sqrt{x-1}-1$$

$$1-\sqrt{x+1}$$

$$\lim_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-3}$$  

$$\lim_{x\rightarrow1}\frac{x-1}{1-\sqrt{x}}$$  

$$\lim_{x\rightarrow\infty}\frac{x-1}{\sqrt{x}-1}$$  

$$\lim_{x\rightarrow3}\frac{1-\sqrt{x-2}}{3-x}$$  

$$x^5$$  

$$3x^2$$  

$$4x^5$$  

$$7$$  

$$\lim_{x\rightarrow\infty}\frac{2x^2+5x-1}{5x^3-2x^2}$$  

$$\lim_{x\rightarrow\infty}\frac{x^2-1}{4-x^2}$$  

$$\lim_{x\rightarrow\infty}\frac{x-1}{\sqrt{x}-1}$$  

$$\lim_{x\rightarrow1}\frac{x-1}{\sqrt{x}-1}$$  

$$\lim_{x\rightarrow\infty}\frac{2x^2+5x-1}{5x^3-2x^2}=\lim_{x\rightarrow\infty}\frac{2x^2}{5x^3}$$

$$\lim_{x\rightarrow\infty}\frac{2x^2+5x-1}{5x^3-2x^2}=\lim_{x\rightarrow\infty}\frac{x^2}{x^3}$$

$$\lim_{x\rightarrow\infty}\frac{2x^2+5x-1}{5x^3-2x^2}=\lim_{x\rightarrow\infty}\frac{2x^2}{-2x^2}$$  

$$\lim_{x\rightarrow\infty}\frac{2x^2+5x-1}{5x^3-2x^2}=\lim_{x\rightarrow\infty}\frac{5x}{5x^3}$$  

$$\infty$$  describes increasing values without bound

$$\infty$$  is not a number

$$\lim_{A\rightarrow\infty}\ \frac{1}{A}=0$$  

$$\lim_{A\rightarrow\infty}\ \frac{1}{A}=\frac{1}{\infty}$$