Question-150560

Question Number 150560 by DELETED last updated on 13/Aug/21

Answered by DELETED last updated on 13/Aug/21

lim_(x→3) (((x^2 −7x+12)sin (x−3))/((x^2 −x−6)^2 )) lim_(x→3) (( (x−4)(x−3)sin (x−3))/({(x+2)(x−3)}^2 )) lim_(x→3) (( (x−4))/((x+2)^2 (x−3)))×lim_((x−3)→0) ((sin (x−3))/((x−3))) (((3−4))/((3+2)^2 (3−4))) × 1=(( (−1)×1)/(5^2 (−1)))=((−1)/(−25))=(1/(25))

$$\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{7x}+\mathrm{12}\right)\mathrm{sin}\:\left(\mathrm{x}−\mathrm{3}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} }\:\:\:\:\:\:\: \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\:\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}−\mathrm{3}\right)\mathrm{sin}\:\left(\mathrm{x}−\mathrm{3}\right)}{\left\{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)\right\}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\:\left(\mathrm{x}−\mathrm{4}\right)}{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{x}−\mathrm{3}\right)}×\underset{\left({x}−\mathrm{3}\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{x}−\mathrm{3}\right)}{\left(\mathrm{x}−\mathrm{3}\right)} \\ $$$$\frac{\left(\mathrm{3}−\mathrm{4}\right)}{\left(\mathrm{3}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{3}−\mathrm{4}\right)}\:×\:\mathrm{1}=\frac{\:\left(−\mathrm{1}\right)×\mathrm{1}}{\mathrm{5}^{\mathrm{2}} \left(−\mathrm{1}\right)}=\frac{−\mathrm{1}}{−\mathrm{25}}=\frac{\mathrm{1}}{\mathrm{25}}\: \\ $$$$ \\ $$

Commented by tabata last updated on 13/Aug/21

$${i}\:{thing}\:=−\frac{\mathrm{1}}{\mathrm{25}} \\ $$

Answered by DELETED last updated on 13/Aug/21

2). lim_(x→−2) ((tan (6x+12))/((4x+8)))= lim_((x+2)→0) ((tan 6(x+2))/(4(x+2))) (6/4) ×lim_((x+2)→0) ((tan (x+2))/((x+2)))=(6/4)×1=(3/2)

$$\left.\mathrm{2}\right).\:\underset{{x}\rightarrow−\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{6x}+\mathrm{12}\right)}{\left(\mathrm{4x}+\mathrm{8}\right)}=\:\underset{\left({x}+\mathrm{2}\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{6}\left(\mathrm{x}+\mathrm{2}\right)}{\mathrm{4}\left(\mathrm{x}+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{6}}{\mathrm{4}}\:×\underset{\left({x}+\mathrm{2}\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(\mathrm{x}+\mathrm{2}\right)}{\left(\mathrm{x}+\mathrm{2}\right)}=\frac{\mathrm{6}}{\mathrm{4}}×\mathrm{1}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by DELETED last updated on 13/Aug/21

3). lim_((x−(π/8) )→0) ((tan (6x−((3π)/4)))/(tan (2x−(π/4)))) = lim_((x−(π/8)) → 0) ((tan 6(x−(π/8)))/(tan 2(x−(π/8)))) = (6/2)×lim_((x−(π/8)) →0) ((tan (x−(π/8)))/(tan (x−(π/8)))) =(6/2) ×1=3 //

$$\left.\mathrm{3}\right).\:\underset{\left({x}−\frac{\pi}{\mathrm{8}}\:\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{tan}\:\left(\mathrm{6x}−\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}{\mathrm{tan}\:\left(\mathrm{2x}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\:\:\:\:\:\:\:=\:\underset{\left({x}−\frac{\pi}{\mathrm{8}}\right)\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{tan}\:\mathrm{6}\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)}{\mathrm{tan}\:\mathrm{2}\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{6}}{\mathrm{2}}×\underset{\left({x}−\frac{\pi}{\mathrm{8}}\right)\:\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{tan}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)}{\mathrm{tan}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{6}}{\mathrm{2}}\:×\mathrm{1}=\mathrm{3}\:// \\ $$

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