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Question-137157




Question Number 137157 by mathlove last updated on 30/Mar/21
Commented by mathlove last updated on 30/Mar/21
with out macloreen sirees and H-pital rools
$${with}\:{out}\:{macloreen}\:{sirees}\:{and}\:{H}-{pital}\:{rools} \\ $$
Commented by mathlove last updated on 30/Mar/21
Commented by bobhans last updated on 30/Mar/21
L′Hopital  lim_(x→0)  ((1−cos x)/(1−sec^2 x)) = lim_(x→0)  ((cos^2 x(1−cos x))/((cos x−1)(cos x+1)))  = −(1/2)
$$\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)\left(\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mathlove last updated on 30/Mar/21
weth out   L  Hopital  rols
$${weth}\:{out}\:\:\:{L}\:\:{Hopital}\:\:{rols} \\ $$
Commented by bobhans last updated on 30/Mar/21
 lim_(x→0)  ((x−sin x)/x^3 ) . lim_(x→0)  (x^3 /(x−tan x))  you can solve it without L′Hopital
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}−\mathrm{tan}\:\mathrm{x}} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{without}\:\mathrm{L}'\mathrm{Hopital} \\ $$
Commented by mathlove last updated on 30/Mar/21
no  I  can not
$${no}\:\:{I}\:\:{can}\:{not} \\ $$

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