Question 11 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3-3\text{x}+1}{\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3-3\text{x}+1}{\text{x}-1}=\frac{(-1)^3-3(-1)+1}{(-1-1)}$
$=\frac{-1+3+1}{-2}=\frac{3}{-2}=\frac{-3}{2}$
View full question & answer→Question 21 Mark
Evaluate the following one sided limits:
$\lim\limits_{\text{x}\rightarrow\frac{\pi^-}{2}}\tan\text{x}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi^-}{2}}\tan\text{ x}$
$=\lim\limits_{\text{h}\rightarrow0}\tan\big(\frac{\pi}{2}-\text{h}\big)$
$=\tan\big(\frac{\pi}{2}-0\big)$
$\Rightarrow\tan\frac{\pi}{2}=\infty$
View full question & answer→Question 31 Mark
Evaluate the following limits:
$\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}^2+3\text{x}+4}{\text{x}^2+3\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}^2+3\text{x}+4}{\text{x}^2+3\text{x}+2}=\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}^2+3\text{x}+4}{(\text{x}+2)(\text{x}+1)}$
$=\frac{2(0)+3(0)+4}{(0+2)(0+1)}=\frac{4}{2}=2$
View full question & answer→Question 41 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{{\text{x}+\text{a}}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sqrt{\text{x}}+\sqrt{\text{a}}}{{\text{x}+\text{a}}}=\frac{\sqrt{\text{a}}+\sqrt{\text{a}}}{\text{a}+\text{a}}$
$=\frac{2\sqrt{\text{a}}}{2\text{a}}=\frac{1}{\sqrt{\text{a}}}$
View full question & answer→Question 51 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{1+(\text{x}-1)^2}{1+{\text{x}^2}}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{1+({\text{x}}-1)}{1+{\text{x}^2}}=\frac{1+0^2}{1+1}=\frac{1}{2}$
View full question & answer→Question 61 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^2-16}{\sqrt{\text{x}}-2}$
Answer$\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^2-16}{\sqrt{\text{x}}-2}$
$=\lim\limits_{\text{x}\rightarrow{4}}\frac{(\text{x}-4)(\text{x}+4)}{\big(\sqrt{\text{x}}-2\big)}$
$=\lim\limits_{\text{x}\rightarrow{4}}\frac{\big(\sqrt{\text{x}}-2\big)\big(\sqrt{\text{x}}+2\big)(\text{x}+4)}{\big(\sqrt{\text{x}}-2\big)}$
$=\lim\limits_{\text{x}\rightarrow4}{\Big(\sqrt{\text{x}}+2\Big)}{(\text{x}+4)}$
$=(2+2)(4+4)$
$=4(8)$
$=32$
View full question & answer→Question 71 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^3-125}{{\text{x}^2-7\text{x}+10}}$
Answer$\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^3-125}{{\text{x}^2-7\text{x}+10}}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\text{x}^2+25+5\text{x}\big)}{(\text{x}-2)(\text{x}-5)}$
$=\frac{(5)^2+25+5(5)}{(5-2)}$
$=\frac{25+25+25}{3}$
$=\frac{75}{3}=25$
View full question & answer→Question 81 Mark
Evaluate the following limits:
$\lim\limits_{\text{x}\rightarrow3}\frac{\sqrt{2\text{x}+3}}{\text{x}+3}$
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\sqrt{2\text{x}+3}}{\text{x}+3}=\frac{\sqrt{9}}{6}=\frac12$
View full question & answer→Question 91 Mark
Find $\lim\limits_{\text{x}\rightarrow1^+}\frac{1}{\text{x}-1}.$
Answer$\lim\limits_{\text{x}\rightarrow1^+}\frac{1}{(\text{x}-1)}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{(1+\text{h}-1)}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}=\infty$
View full question & answer→Question 101 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{1}{4}}\frac{4\text{x}-1}{2\sqrt{\text{x}}-1}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{1}{4}}\frac{4\text{x}-1}{2\sqrt{\text{x}}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{1}{4}}\frac{4\Big(\text{x}-\frac14\Big)}{2\Big(\sqrt{\text{x}}-\frac12\Big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{1}{4}}\frac{\Big(\sqrt{\text{x}}-\frac12\Big)\Big(\sqrt{\text{x}}+\frac12\Big)}{2\Big(\sqrt{\text{x}}-\frac12\Big)}$
$=\lim\limits_{\text{x}\rightarrow\frac14}\frac{\Big(\sqrt{\text{x}}+\frac12\Big)}{2}$
$=\frac{4\big(\frac12+\frac12\big)}{2}$
$=\frac{4(1)}{2}=2$
View full question & answer→Question 111 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-5}\frac{2\text{x}^2+9\text{x}-5}{\text{x}+5}$
Answer$\lim\limits_{\text{x}\rightarrow-5}\frac{2\text{x}^2+9\text{x}-5}{\text{x}+5}=\lim\limits_{\text{x}\rightarrow-5}\frac{(\text{x}+5)(2\text{x}-1)}{(\text{x}+5)}$
$=\lim\limits_{\text{x}\rightarrow-5}(2\text{x}-1)=2(-5)-1=-10-1=-11$
View full question & answer→Question 121 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^4-81}{\text{x}^2-9}$
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^4-81}{\text{x}^2-9}$$=\lim\limits_{\text{x}\rightarrow3}\frac{\big(\text{x}^2-9\big)\big(\text{x}^2+9\big)}{\big(\text{x}^2-9\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\text{x}^2+9=(3)^2+9=9+9=18$
View full question & answer→Question 131 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x}+8}}{\sqrt{\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x}+8}}{\sqrt{\text{x}}}=\frac{\sqrt{(1+8)}}{\sqrt{1}}=\sqrt{9}=3$
View full question & answer→Question 141 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\text{x}+1}{\text{x}+3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\text{x}+1}{\text{x}+3}=\frac{3(0)+1}{(0+3)}=\frac13$
View full question & answer→Question 151 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a}+\text{x})^2-\text{a}^2}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a}+\text{x})^2-\text{a}^2}{\text{x}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a}+\text{x}-\text{a})(\text{a}+\text{x}+\text{a})}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{x})(2\text{a}+\text{x})}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}(2\text{a}+\text{x})$
$=2\text{a}$
View full question & answer→Question 161 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}9$
Answer$\lim\limits_{\text{x}\rightarrow0}9=9$
View full question & answer→Question 171 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^{\frac{2}{3}}-9}{{\text{x}}-27}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{{\text{x}^{\frac{2}{3}}}-9}{{\text{x}-27}}=\frac{-9}{-27}=\frac{1}{3}$
View full question & answer→Question 181 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}(3-\text{x})$
Answer$\lim\limits_{\text{x}\rightarrow2}(3-\text{x})=(3-2)=1$
View full question & answer→Question 191 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-9}{\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-9}{\text{x}+2}=\frac{3^2-9}{3+2}=0$
View full question & answer→Question 201 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-8}{\text{x}^2-4}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-8}{\text{x}^2-4}$$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)\big(\text{x}^2+4+2\text{x}\big)}{(\text{x}-2)(\text{x}+2)}$
$=\frac{(2)^2+4+2(2)}{2+2}=\frac{4+4+4}{4}=\frac{12}{4}=3$
View full question & answer→Question 211 Mark
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\Big(1+\frac{\text{x}}{\text{n}}\Big)^\text{n}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\Big(1+\frac{\text{x}}{\text{n}}\Big)^\text{n}$
$=\text{e}^{\lim\limits_{\text{n} \rightarrow\infty}\big(\frac{\text{x}}{\text{n}}\big)\text{n}}$
$=\text{e}^\text{x}$
View full question & answer→Question 221 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-1}\big(4\text{x}^2+2\big)$
Answer$\lim\limits_{\text{x}\rightarrow-1}\big(4\text{x}^2+2\big)=\big(4(-1)^2+2\big)=6$
View full question & answer→Question 231 Mark
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{b}}{(\text{x}+\text{d})}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{b}}{(\text{x}+\text{d})}=\frac{\text{a}\times0+\text{b}}{(0+\text{d})}=\frac{\text{b}}{\text{d}}$
View full question & answer→Question 241 Mark
Evaluate the following one sided limits:
$\lim\limits_{\text{x}\rightarrow0^+}\frac{2}{\text{x}^{\frac{1}{5}}}.$
Answer$\lim\limits_{\text{x}\rightarrow0^+}\frac{2}{\text{x}^{\frac{1}{5}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2}{(0+\text{h})^{\frac{1}{5}}}$
$\Rightarrow\frac{2}{0}=\infty$
View full question & answer→Question 251 Mark
Evaluate the following limits:
$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2+1}{\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2+1}{\text{x}+1}=\frac{(1)^2+1}{1+1}=\frac{1+1}{2}=\frac{2}{2}=1$
View full question & answer→