Menu Close

lim-x-0-1-6x-2-1-7x-x-2-x-3-




Question Number 144816 by liberty last updated on 29/Jun/21
 lim_(x→0)  (((√(1+6x^2 ))−(1+7x))/(x^2 (x−3))) =?
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\mathrm{6x}^{\mathrm{2}} }−\left(\mathrm{1}+\mathrm{7x}\right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}−\mathrm{3}\right)}\:=? \\ $$
Answered by imjagoll last updated on 29/Jun/21
 =−(1/3)lim_(x→0)  (((1+6x^2 )−(1+14x+49x^2 ))/(2x^2 ))  = −(1/6)lim_(x→0)  ((−43x^2 −14x)/x^2 )  = ∞
$$\:=−\frac{\mathrm{1}}{\mathrm{3}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{6x}^{\mathrm{2}} \right)−\left(\mathrm{1}+\mathrm{14x}+\mathrm{49x}^{\mathrm{2}} \right)}{\mathrm{2x}^{\mathrm{2}} } \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{6}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{43x}^{\mathrm{2}} −\mathrm{14x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\:\infty\: \\ $$
Answered by mathmax by abdo last updated on 29/Jun/21
f(x)=(((√(1+6x^2 ))−1−7x)/(x^2 (x−3))) ⇒f(x)∼((1+3x^2 −1−7x)/(x^2 (x−3)))  ⇒f(x)∼(((3x−7))/(x(x−3))) ⇒lim_(x→0)   f(x)=∞
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{1}+\mathrm{6x}^{\mathrm{2}} }−\mathrm{1}−\mathrm{7x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}−\mathrm{3}\right)}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1}+\mathrm{3x}^{\mathrm{2}} −\mathrm{1}−\mathrm{7x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}−\mathrm{3}\right)} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\left(\mathrm{3x}−\mathrm{7}\right)}{\mathrm{x}\left(\mathrm{x}−\mathrm{3}\right)}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\mathrm{f}\left(\mathrm{x}\right)=\infty \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *