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Question Number 72688 by Rio Michael last updated on 31/Oct/19

Evaluate     lim_(t→9) ((9−t)/(3−(√t) ))

$${Evaluate}\: \\ $$$$\:\underset{{t}\rightarrow\mathrm{9}} {\:{lim}}\frac{\mathrm{9}−{t}}{\mathrm{3}−\sqrt{{t}}\:} \\ $$

Answered by JDamian last updated on 01/Nov/19

lim_(t→9) (((9−t)/(3−(√t) ))×((3+(√t))/(3+(√t))))=lim_(t→9) (((9−t)(3+(√t)))/(9−t))=  =lim_(t→9)  (3+(√t))=6

$$\underset{{t}\rightarrow\mathrm{9}} {\mathrm{lim}}\left(\frac{\mathrm{9}−{t}}{\mathrm{3}−\sqrt{{t}}\:}×\frac{\mathrm{3}+\sqrt{{t}}}{\mathrm{3}+\sqrt{{t}}}\right)=\underset{{t}\rightarrow\mathrm{9}} {\mathrm{lim}}\frac{\left(\mathrm{9}−{t}\right)\left(\mathrm{3}+\sqrt{{t}}\right)}{\mathrm{9}−{t}}= \\ $$$$=\underset{{t}\rightarrow\mathrm{9}} {\mathrm{lim}}\:\left(\mathrm{3}+\sqrt{{t}}\right)=\mathrm{6} \\ $$

Commented by Rio Michael last updated on 31/Oct/19

thanks sir

$${thanks}\:{sir} \\ $$

Answered by malwaan last updated on 31/Oct/19

put t=x^(2 )    ⇒lim_(x→3)  ((9−x^2 )/(3−x))   = lim_(x→3)  (((3+x)(3−x))/(3−x))  = lim_(x→3) (3+x)=3+3=6

$$\boldsymbol{{put}}\:\boldsymbol{{t}}=\boldsymbol{{x}}^{\mathrm{2}\:} \: \\ $$$$\Rightarrow\underset{\boldsymbol{{x}}\rightarrow\mathrm{3}} {\boldsymbol{{lim}}}\:\frac{\mathrm{9}−\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{3}−\boldsymbol{{x}}} \\ $$$$\:=\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{3}} {\boldsymbol{{lim}}}\:\frac{\left(\mathrm{3}+\boldsymbol{{x}}\right)\left(\mathrm{3}−\boldsymbol{{x}}\right)}{\mathrm{3}−\boldsymbol{{x}}} \\ $$$$=\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{3}} {\boldsymbol{{lim}}}\left(\mathrm{3}+\boldsymbol{{x}}\right)=\mathrm{3}+\mathrm{3}=\mathrm{6} \\ $$

Answered by petrochengula last updated on 01/Nov/19

=lim_(t→9) (((3−(√t))(3+(√t)))/(3−(√t)))=lim_(t→9) (3+(√t))=6

$$=\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{9}} \frac{\left(\mathrm{3}−\sqrt{\mathrm{t}}\right)\left(\mathrm{3}+\sqrt{\mathrm{t}}\right)}{\mathrm{3}−\sqrt{\mathrm{t}}}=\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{9}} \left(\mathrm{3}+\sqrt{\mathrm{t}}\right)=\mathrm{6} \\ $$

Commented by Rio Michael last updated on 01/Nov/19

thanks

$${thanks} \\ $$

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