Question Number 80300 by john santu last updated on 02/Feb/20
Commented by ~blr237~ last updated on 02/Feb/20
![lim_(x→−∞) (1/x)=0^(− ) and lim_(x→0^− ) (1/x)=−∞ lim_(x→0^− ) 3x+5x^2 = 0^− due to the sign of x(3+5x)<0 when x∈]−(3/5),0[ samely when x→0^− so the ans lim_(x→−∞) e^((6x^2 +x)/(5+3x)) = 0](Q80304.png)
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\:\frac{\mathrm{1}}{{x}}=\mathrm{0}^{−\:\:} \:\:{and}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\:\frac{\mathrm{1}}{{x}}=−\infty\:\: \\ $$$$\left.\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\:\mathrm{3}{x}+\mathrm{5}{x}^{\mathrm{2}} =\:\mathrm{0}^{−} \:\:{due}\:{to}\:{the}\:{sign}\:{of}\:\:{x}\left(\mathrm{3}+\mathrm{5}{x}\right)<\mathrm{0}\:{when}\:{x}\in\right]−\frac{\mathrm{3}}{\mathrm{5}},\mathrm{0}\left[\:\:{samely}\:{when}\:{x}\rightarrow\mathrm{0}^{−} \:\right. \\ $$$${so}\:{the}\:{ans}\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\:{e}^{\frac{\mathrm{6}{x}^{\mathrm{2}} +{x}}{\mathrm{5}+\mathrm{3}{x}}} =\:\mathrm{0} \\ $$$$ \\ $$
Commented by john santu last updated on 02/Feb/20
$${what}\:{wrong}\:{this}\:{method}?\:{i}\:{see}\:{in}\:{book} \\ $$$${calculus}\:\mathrm{9}^{{th}} \:{edition}\:{by}\:{Purcel}\:{like} \\ $$$${this} \\ $$
Commented by ~blr237~ last updated on 02/Feb/20
![mean that the book assumed that you understood all the parts i explain secondly don′t continue to use ((1/0)=∞) it′s not appropriate But (1/0^+ )=+∞ and (1/0^− ) =−∞ x→0^+ ⇔ x∈]0,a[ with a→0 x→0^− ⇔ x∈]a,0[ with a →0](Q80311.png)
$${mean}\:{that}\:{the}\:{book}\:{assumed}\:{that}\:{you}\:{understood}\:{all}\:{the}\:{parts}\:{i}\:{explain}\: \\ $$$${secondly}\:\:{don}'{t}\:{continue}\:{to}\:{use}\:\:\left(\frac{\mathrm{1}}{\mathrm{0}}=\infty\right)\:{it}'{s}\:\:{not}\:{appropriate} \\ $$$${But}\:\:\frac{\mathrm{1}}{\mathrm{0}^{+} }=+\infty\:{and}\:\frac{\mathrm{1}}{\mathrm{0}^{−} }\:=−\infty\: \\ $$$$\left.{x}\rightarrow\mathrm{0}^{+} \:\:\Leftrightarrow\:{x}\in\right]\mathrm{0},{a}\left[\:{with}\:{a}\rightarrow\mathrm{0}\right. \\ $$$$\left.{x}\rightarrow\mathrm{0}^{−} \:\Leftrightarrow\:{x}\in\right]{a},\mathrm{0}\left[\:\:{with}\:\:{a}\:\rightarrow\mathrm{0}\right. \\ $$$$ \\ $$
Commented by MJS last updated on 02/Feb/20
$$\frac{\mathrm{6}{x}^{\mathrm{2}} +{x}}{\mathrm{3}{x}+\mathrm{5}}=\mathrm{2}{x}−\mathrm{3}+\frac{\mathrm{15}}{\mathrm{3}{x}+\mathrm{5}} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\mathrm{2}{x}−\mathrm{3}\right)\:=−\infty;\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{15}}{\mathrm{3}{x}+\mathrm{5}}\:=\mathrm{0} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{6}{x}^{\mathrm{2}} +{x}}{\mathrm{3}{x}+\mathrm{5}}\:=−\infty \\ $$$$\Rightarrow\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\mathrm{e}^{\frac{\mathrm{6}{x}^{\mathrm{2}} +{x}}{\mathrm{3}{x}+\mathrm{5}}} \:=\mathrm{0} \\ $$