Calculus – Limits and continuity – Examples

1) \(\displaystyle\lim\limits_{x\to 0}\frac{x^2+2x+4}{x+2}=\frac{0+0+4}{0+2}=2\).	2) \(\displaystyle\lim\limits_{x\to 1}\frac{x^2-1}{x+1}=\frac{0}{2}=0\).
3) \(\displaystyle\lim\limits_{x\to 1}\frac{x^2-1}{x-1}=\lim\limits_{x\to 1}\frac{(x-1)(x+1)}{x-1}=\lim\limits_{x\to 1}\frac{x+1}{1}=2\).	4) \(\displaystyle\lim\limits_{x\to -3}\frac{x^2+5x+6}{x+3}=\lim\limits_{x\to -3}\frac{(x+2)(x+3)}{x+3}=\lim\limits_{x\to -3}\frac{x+2}{1}=-1\).
5) \(\displaystyle\lim\limits_{x\to 3}\frac{x^2-3x}{x^2-9}=\lim\limits_{x\to 3}\frac{x(x-3)}{(x-3)(x+3)}=\lim\limits_{x\to 3}\frac{x}{x+3}=\frac{1}{2}\).	6) \(\displaystyle\lim\limits_{x\to -2}\frac{x^2+4x+4}{x^2+6x+8}=\lim\limits_{x\to -2}\frac{(x+2)^2}{(x+2)(x+4)}=\lim\limits_{x\to -2}\frac{x+2}{x+4}=\frac{0}{4}=0\).
7) \(\displaystyle\lim\limits_{h\to 0}\frac{(h+4)^3-64}{h}=\lim\limits_{h\to 0}\frac{h^3+12h^2+48h+64-64}{h}=\lim\limits_{h\to 0}(h^2+12h+48)=48\).
8) \(\displaystyle\lim\limits_{x\to 0}\frac{\sin(2x)}{\sin(x)}=\lim\limits_{x\to 0}\frac{2\sin(x)\cos(x)}{\sin(x)}=\lim\limits_{x\to 0}\frac{2\cos(x)}{1}=2\cos(0)=2\).
9) \(\displaystyle\lim\limits_{x\to 0}\frac{\tan(x)}{\cos(x)}=\lim\limits_{x\to 0}\frac{\tan(0)}{\cos(0)}=\frac{0}{1}=0\).	10) \(\displaystyle\lim\limits_{x\to 0}\frac{\tan(x)}{\sin(x)}=\lim\limits_{x\to 0}\frac{\sin(x)}{\cos(x)}\cdot\frac{1}{\sin(x)}=\lim\limits_{x\to 0}\frac{1}{\cos(x)}=\frac{1}{\cos(0)}=1\).
11) \(\displaystyle\lim\limits_{x\to 0}x\sin\left(\frac{1}{x}\right)=0\) by the squeeze theorem, since \(-x\displaystyle\leq x\sin\left(\frac{1}{x}\right)\leq x\) and \(\lim\limits_{x\to 0}(-x)=0=\lim\limits_{x\to 0}x\).
12) \(\displaystyle\lim\limits_{x\to 5}\frac{\displaystyle\frac{1}{x}-\frac{1}{5}}{x-5}=\lim\limits_{x\to 5}\frac{\displaystyle\frac{5-x}{5x}}{x-5}=\lim\limits_{x\to 5}\frac{-1}{5x}=-\frac{1}{25}\).	13) \(\displaystyle\lim\limits_{t\to 0}\left(\frac{1}{t(t+1)}-\frac{1}{t}\right)=\lim\limits_{t\to 0}\frac{1-(t+1)}{t(t+1)}=\lim\limits_{t\to 0}\frac{-1}{t+1}=-1\).
14) \(\displaystyle\lim\limits_{t\to 0}\frac{\sqrt{4+t}-2}{t}=\lim\limits_{t\to 0}\frac{\sqrt{4+t}-2}{t}\cdot\frac{\sqrt{4+t}+2}{\sqrt{4+t}+2}=\lim\limits_{t\to 0}\frac{4+t-4}{t\left(\sqrt{4+t}+2\right)}=\lim\limits_{t\to 0}\frac{1}{\sqrt{4+t}+2}=\frac{1}{2+2}=\frac{1}{4}\).
15) \(\displaystyle\lim\limits_{t\to 0}\frac{\sqrt{1+t}-\sqrt{1-t}}{t}=\lim\limits_{t\to 0}\frac{\sqrt{1+t}-\sqrt{1-t}}{t}\cdot\frac{\sqrt{1+t}+\sqrt{1-t}}{\sqrt{1+t}+\sqrt{1-t}}=\lim\limits_{t\to 0}\frac{1+t-(1-t)}{t\left(\sqrt{1+t}+\sqrt{1-t}\right)}=\lim\limits_{t\to 0}\frac{2}{\sqrt{1+t}+\sqrt{1-t}}=1\).
16) \(\displaystyle\lim\limits_{t\to 1}\frac{t^5-1}{t^3-1}=\lim\limits_{t\to 1}\frac{(t-1)(t^4+t^3+t^2+t+1)}{(t-1)(t^2+t+1)}=\lim\limits_{t\to 1}\frac{t^4+t^3+t^2+t+1}{t^2+t+1}=\frac{5}{3}\).
17) \(\displaystyle\lim\limits_{x\uparrow\frac{1}{2}\pi}\tan(x)=\infty\).	18) \(\displaystyle\lim\limits_{x\downarrow\frac{1}{2}\pi}\tan(x)=-\infty\).
19) \(\displaystyle\lim\limits_{x\to 1}e^{\displaystyle\frac{x^2-2x+1}{x^2-3x+2}}=e^{\displaystyle\lim\limits_{x\to 1}\frac{(x-1)^2}{(x-1)(x-2)}}=e^0=1\).	20) \(\displaystyle\lim\limits_{x\downarrow 1}\ln\left(\frac{1-x}{1-\sqrt{x}}\right)=\ln\left(\lim\limits_{x\downarrow 1}\frac{(1-\sqrt{x})(1+\sqrt{x})}{1-\sqrt{x}}\right)=\ln(2)\).
21) \(\displaystyle\lim\limits_{x\to 1}\arcsin\left(\frac{1-x}{1-x^2}\right)=\arcsin\left(\lim\limits_{x\to 1}\frac{1-x}{(1-x)(1+x)}\right)=\arcsin\left(\frac{1}{2}\right)=\frac{1}{6}\pi\).
22) \(\displaystyle\lim\limits_{x\to 0}\arctan\left(\frac{1+e^x}{1+e^{2x}}\right)=\arctan\left(\lim\limits_{x\to 0}\frac{1+e^x}{1+e^{2x}}\right)=\arctan\left(\frac{1+1}{1+1}\right)=\arctan(1)=\frac{1}{4}\pi\).

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Last modified on October 26, 2021