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Question Number 134746 by zahaku last updated on 06/Mar/21
How to calculate  lim_(x→−3) ((x+3)/( (√(∣x^2 +x−6∣))))    ?
$${How}\:{to}\:{calculate} \\ $$$${li}\underset{{x}\rightarrow−\mathrm{3}} {{m}}\frac{{x}+\mathrm{3}}{\:\sqrt{\mid{x}^{\mathrm{2}} +{x}−\mathrm{6}\mid}}\:\:\:\:? \\ $$
Answered by mathmax by abdo last updated on 06/Mar/21
0
$$\mathrm{0} \\ $$
Answered by EDWIN88 last updated on 07/Mar/21
lim_(x→−3)  (((x+3))/( (√(∣(x+3)(x−2)∣)))) = lim_(x→−3)  (1/( (√(2−x)))) .lim_(x→−3)  (((x+3))/( (((x+3)^2 ))^(1/4) ))  = (1/( (√5))).lim_(x→−3) (√(x+3)) = 0
$$\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}+\mathrm{3}\right)}{\:\sqrt{\mid\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}−\mathrm{2}\right)\mid}}\:=\:\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−\mathrm{x}}}\:.\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}+\mathrm{3}\right)}{\:\sqrt[{\mathrm{4}}]{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}.\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}}\sqrt{\mathrm{x}+\mathrm{3}}\:=\:\mathrm{0} \\ $$$$ \\ $$

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