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Question Number 106825 by 175mohamed last updated on 07/Aug/20
lim_(x→0) ((sin5x − tan5x)/x^3 )
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\mathrm{5}{x}\:−\:{tan}\mathrm{5}{x}}{{x}^{\mathrm{3}} } \\ $$
Answered by bemath last updated on 07/Aug/20
^(@bemath@) lim_(x→0) ((tan 5x(cos 5x−1))/x^3 )= lim_(x→0) ((tan 5x(−2sin^2 (((5x)/2))))/x^3 )=−2×5×((25)/4) =−((125)/2)
$$\:\:\:\:\:\:\:^{@\mathrm{bemath}@} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{5x}\left(\mathrm{cos}\:\mathrm{5x}−\mathrm{1}\right)}{\mathrm{x}^{\mathrm{3}} }= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{5x}\left(−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{5x}}{\mathrm{2}}\right)\right)}{\mathrm{x}^{\mathrm{3}} }=−\mathrm{2}×\mathrm{5}×\frac{\mathrm{25}}{\mathrm{4}} \\ $$$$=−\frac{\mathrm{125}}{\mathrm{2}} \\ $$
Answered by Dwaipayan Shikari last updated on 07/Aug/20
lim_(x→0) ((sin5x(cos5x−1))/(cos5x.x^3 ))=lim_(x→0) ((−5x.2sin^2 (((5x)/2)))/x^3 )=lim_(x→0) ((−5x.2(((5x)/2))^2 )/x^3 )=−((125)/2)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\mathrm{5}{x}\left({cos}\mathrm{5}{x}−\mathrm{1}\right)}{{cos}\mathrm{5}{x}.{x}^{\mathrm{3}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{5}{x}.\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{5}{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{3}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{5}{x}.\mathrm{2}\left(\frac{\mathrm{5}{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{{x}^{\mathrm{3}} }=−\frac{\mathrm{125}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 07/Aug/20
Another way lim_(x→0) ((5x−(((5x)^3 )/6)−5x−(((5x)^3 )/3))/x^3 )=lim_(x→0) ((−(((375x^3 )/6)))/x^3 )=−((125)/2)
$${Another}\:{way} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{5}{x}−\frac{\left(\mathrm{5}{x}\right)^{\mathrm{3}} }{\mathrm{6}}−\mathrm{5}{x}−\frac{\left(\mathrm{5}{x}\right)^{\mathrm{3}} }{\mathrm{3}}}{{x}^{\mathrm{3}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\left(\frac{\mathrm{375}{x}^{\mathrm{3}} }{\mathrm{6}}\right)}{{x}^{\mathrm{3}} }=−\frac{\mathrm{125}}{\mathrm{2}} \\ $$
Answered by malwaan last updated on 07/Aug/20
lim_(x→0) ((sin 5x −((sin 5x)/(cos 5x)))/x^3 ) = lim_(x→0) ((sin5x cos5x−sin5x)/(x^3 cos5x)) =lim_(x→0) ((sin5x[cos2(((5x)/2))−1])/(x^3 cos5x)) =lim_(x→0) ((sin5x[−2sin^2 (((5x)/2))])/(x^3 cos5x)) = ((5[−2((5/2))^2 ])/(1^3 ×1)) = −((125)/2)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\:\mathrm{5}{x}\:−\frac{{sin}\:\mathrm{5}{x}}{{cos}\:\mathrm{5}{x}}}{{x}^{\mathrm{3}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{sin}\mathrm{5}{x}\:{cos}\mathrm{5}{x}−{sin}\mathrm{5}{x}}{{x}^{\mathrm{3}} {cos}\mathrm{5}{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\mathrm{5}{x}\left[{cos}\mathrm{2}\left(\frac{\mathrm{5}{x}}{\mathrm{2}}\right)−\mathrm{1}\right]}{{x}^{\mathrm{3}} {cos}\mathrm{5}{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\mathrm{5}{x}\left[−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{5}{x}}{\mathrm{2}}\right)\right]}{{x}^{\mathrm{3}} {cos}\mathrm{5}{x}} \\ $$$$=\:\frac{\mathrm{5}\left[−\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} \right]}{\mathrm{1}^{\mathrm{3}} ×\mathrm{1}}\:=\:−\frac{\mathrm{125}}{\mathrm{2}} \\ $$
Commented by JDamian last updated on 07/Aug/20
Why x^3 ⇒ 1^3 ?
$${Why}\:\:\:\:{x}^{\mathrm{3}} \:\Rightarrow\:\mathrm{1}^{\mathrm{3}} \:? \\ $$
Commented by malwaan last updated on 07/Aug/20
sir JDamian it is shortcut method with sinx/tanx/x
$${sir}\:{JDamian} \\ $$$${it}\:{is}\:{shortcut}\:{method}\:{with} \\ $$$${sinx}/{tanx}/{x}\: \\ $$
Answered by 1549442205PVT last updated on 07/Aug/20
lim_(x→0) ((sin5x − tan5x)/x^3 ) = _(L′Hopital) lim_(x→0) ((5cos5x−(1+tan^2 5x).5)/(3x^2 )) = _(L′Hopital) lim_(x→0) ((−25sin5x−5.[2tan5x(1+tan^2 5x).5])/(6x)) =lim_(x→0) ((−25sin5x−50tan5x−50tan^3 5x)/(6x)) =lim_(x→0) (((−125)/6)×((sin5x)/(5x)))−lim_(x→0) (((250)/(6cos5x))×((sin5x)/(5x))) −lim_(x→0) (((250)/(6cosx))×((sin5x)/(5x))×tan^2 x) =(((−125)/6)×1)−(((250)/6)×1)−(((250)/6)×1×0) =((−375)/6)=((−125)/2)
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\mathrm{5}{x}\:−\:{tan}\mathrm{5}{x}}{{x}^{\mathrm{3}} }\:\:\:\:\:=\underset{\mathrm{L}'\mathrm{Hopital}} {\:}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{5cos5x}−\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{5x}\right).\mathrm{5}}{\mathrm{3x}^{\mathrm{2}} } \\ $$$$\underset{\mathrm{L}'\mathrm{Hopital}} {=\:\:\:}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{25sin5x}−\mathrm{5}.\left[\mathrm{2tan5x}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{5x}\right).\mathrm{5}\right]}{\mathrm{6x}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{25sin5x}−\mathrm{50tan5x}−\mathrm{50tan}^{\mathrm{3}} \mathrm{5x}}{\mathrm{6x}} \\ $$$$=\mathrm{l}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{im}}\left(\frac{−\mathrm{125}}{\mathrm{6}}×\frac{\mathrm{sin5x}}{\mathrm{5x}}\right)−\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{250}}{\mathrm{6cos5x}}×\frac{\mathrm{sin5x}}{\mathrm{5x}}\right) \\ $$$$−\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{250}}{\mathrm{6cosx}}×\frac{\mathrm{sin5x}}{\mathrm{5x}}×\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right) \\ $$$$=\left(\frac{−\mathrm{125}}{\mathrm{6}}×\mathrm{1}\right)−\left(\frac{\mathrm{250}}{\mathrm{6}}×\mathrm{1}\right)−\left(\frac{\mathrm{250}}{\mathrm{6}}×\mathrm{1}×\mathrm{0}\right) \\ $$$$=\frac{−\mathrm{375}}{\mathrm{6}}=\frac{−\mathrm{125}}{\mathrm{2}} \\ $$

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