Menu Close

1-lim-x-pi-4-sin-pi-4-x-tan-x-pi-4-2-lim-x-pi-2-pi-pi-2x-tan-x-pi-2-2-x-pi-cos-2-x-3-lim-x-0-3x-cos-3x-cos-7x-sin-2x-1-tan-2x-1-4-lim-x-0-sin-2

Tinku Tara 0 Comments
Pin




Question Number 111960 by john santu last updated on 05/Sep/20
(1)lim_(x→(π/4)) sin ((π/4)−x).tan (x+(π/4)) ? (2)lim_(x→(π/2)) ((π(π−2x)tan (x−(π/2)))/(2(x−π)cos^2 x)) ? (3)lim_(x→0) ((3x(cos 3x−cos 7x))/( (√(sin 2x+1)) −(√(tan 2x+1)))) ? (4)lim_(x→0) ((sin 2x)/(3−(√(3x+9)))) ?
$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right).\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\:? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\pi\left(\pi−\mathrm{2}{x}\right)\mathrm{tan}\:\left({x}−\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}\left({x}−\pi\right)\mathrm{cos}\:^{\mathrm{2}} {x}}\:? \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{x}\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{7}{x}\right)}{\:\sqrt{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{1}}\:−\sqrt{\mathrm{tan}\:\mathrm{2}{x}+\mathrm{1}}}\:? \\ $$$$\left(\mathrm{4}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{3}−\sqrt{\mathrm{3}{x}+\mathrm{9}}}\:? \\ $$$$ \\ $$
Answered by bobhans last updated on 05/Sep/20
solution: (1) lim_(x→(π/4)) sin (x−(π/4)).tan (x+(π/4))= set x = (π/4)+z ; z→0 lim_(z→0) sin z. tan (z+(π/2)) = lim_(z→0) ((sin z)/(cot (z+(π/2)))) lim_(z→0) ((sin z)/(tan z)) = 1 (2)lim_(x→(π/2)) ((π(π−2x) tan (x−(π/2)))/(2(x−π)cos^2 x)) = set x = (π/2)+s , s→0 lim_(s→0) ((π(π−π−2s).tan s)/(2(s−(π/2))cos^2 (s+(π/2)))) = lim_(s→0) ((−2πs.tan s)/((2s−π)sin^2 s)) = ((−2π)/(−π)) = 2 (3) lim_(x→0) ((3x(cos 3x−cos 7x))/( (√(1+sin 2x)) −(√(1+tan 2x)))) = lim_(x→0) ((3x(1−((9x^2 )/2)−1+((49x^2 )/2)))/((1+((sin 2x)/2))−(1+((tan 2x)/2)))) = lim_(x→0) ((2×3x(20x^2 ))/(tan 2x(cos 2x−1))) = lim_(x→0) ((60x^2 )/(1−2x^2 −1)) = −30 (4) lim_(x→0) ((sin 2x)/(3−(√(3x+9)))) = lim_(x→0) ((sin 2x)/(3−(√(9(1+(x/3)))))) lim_(x→0) ((sin 2x)/(3(1−(√(1+(x/3)))))) = lim_(x→0) ((sin 2x)/(3(1−(1+(x/6))))) = lim_(x→0) ((sin 2x)/(−((x/2)))) = −4
$$\mathrm{solution}:\: \\ $$$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right).\mathrm{tan}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)= \\ $$$$\:\:\:\:\mathrm{set}\:\mathrm{x}\:=\:\frac{\pi}{\mathrm{4}}+\mathrm{z}\:;\:\mathrm{z}\rightarrow\mathrm{0} \\ $$$$\:\:\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{sin}\:\mathrm{z}.\:\mathrm{tan}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{2}}\right)\:=\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{z}}{\mathrm{cot}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{2}}\right)} \\ $$$$\:\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{z}}{\mathrm{tan}\:\mathrm{z}}\:=\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\pi\left(\pi−\mathrm{2x}\right)\:\mathrm{tan}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}\left(\mathrm{x}−\pi\right)\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:= \\ $$$$\:\:\mathrm{set}\:\mathrm{x}\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{s}\:,\:\mathrm{s}\rightarrow\mathrm{0} \\ $$$$\:\:\:\:\underset{\mathrm{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\pi\left(\pi−\pi−\mathrm{2s}\right).\mathrm{tan}\:\mathrm{s}}{\mathrm{2}\left(\mathrm{s}−\frac{\pi}{\mathrm{2}}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{s}+\frac{\pi}{\mathrm{2}}\right)}\:= \\ $$$$\:\:\:\underset{\mathrm{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2}\pi\mathrm{s}.\mathrm{tan}\:\mathrm{s}}{\left(\mathrm{2s}−\pi\right)\mathrm{sin}\:^{\mathrm{2}} \mathrm{s}}\:=\:\frac{−\mathrm{2}\pi}{−\pi}\:=\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3x}\left(\mathrm{cos}\:\mathrm{3x}−\mathrm{cos}\:\mathrm{7x}\right)}{\:\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2x}}\:−\sqrt{\mathrm{1}+\mathrm{tan}\:\mathrm{2x}}}\:= \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3x}\left(\mathrm{1}−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}+\frac{\mathrm{49x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\left(\mathrm{1}+\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{2}}\right)−\left(\mathrm{1}+\frac{\mathrm{tan}\:\mathrm{2x}}{\mathrm{2}}\right)}\:= \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}×\mathrm{3x}\left(\mathrm{20x}^{\mathrm{2}} \right)}{\mathrm{tan}\:\mathrm{2x}\left(\mathrm{cos}\:\mathrm{2x}−\mathrm{1}\right)}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{60x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:=\:−\mathrm{30} \\ $$$$\left(\mathrm{4}\right)\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{3}−\sqrt{\mathrm{3x}+\mathrm{9}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{3}−\sqrt{\mathrm{9}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{3}}\right)}} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{3}\left(\mathrm{1}−\sqrt{\mathrm{1}+\frac{\mathrm{x}}{\mathrm{3}}}\right)}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{3}\left(\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{6}}\right)\right)} \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}}{−\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\:=\:−\mathrm{4}\: \\ $$

Pin

Leave a Reply

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