Menu Close

Question-190253




Question Number 190253 by 073 last updated on 30/Mar/23
Answered by cortano12 last updated on 30/Mar/23
 L=lim_(x→0)  (((tan x+x)(tan x−x))/(x^2  tan^2 x))   L= lim_(x→0)  ((tan x+x)/x) . lim_(x→0)  ((tan x−x)/(x tan^2 x))   L= 2. lim_(x→0)  ((tan x−x)/x^3 )    L= (2/3)
$$\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{tan}\:\mathrm{x}+\mathrm{x}\right)\left(\mathrm{tan}\:\mathrm{x}−\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}+\mathrm{x}}{\mathrm{x}}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\mathrm{L}=\:\mathrm{2}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\: \\ $$$$\:\mathrm{L}=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Leave a Reply

Your email address will not be published.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