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What-are-the-conditions-whereby-the-limit-of-a-function-does-not-exist-at-a-poont-

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Question Number 27727 by NECx last updated on 13/Jan/18
What are the conditions whereby the limit of a function does not exist at a poont?
$${What}\:{are}\:{the}\:{conditions}\:{whereby} \\ $$$${the}\:{limit}\:{of}\:{a}\:{function}\:{does}\:{not} \\ $$$${exist}\:{at}\:{a}\:{poont}? \\ $$
Answered by prakash jain last updated on 13/Jan/18
When LHL is not equal to RHL.
$$\mathrm{When}\:\mathrm{LHL}\:\mathrm{is}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{RHL}. \\ $$
Commented by NECx last updated on 13/Jan/18
in this case lim_(x→0) ((x^2 +3x−4)/(2−(√(x^2 +4)))) how do I find the LHL and RHL to show that the limit does not exist if at all it doesnt.
$${in}\:{this}\:{case}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$ \\ $$$${how}\:{do}\:{I}\:{find}\:{the}\:{LHL}\:{and}\:{RHL} \\ $$$${to}\:{show}\:{that}\:{the}\:{limit}\:{does}\:{not} \\ $$$${exist}\:{if}\:{at}\:{all}\:{it}\:{doesnt}. \\ $$
Commented by prakash jain last updated on 13/Jan/18
((x^2 +3x−4)/(2−(√(x^2 +4)))) in this 2−(√(x^2 +4)) <0 for all x x^2 +3x−4=−4 near x=0 lim_(x→0) ((x^2 +3x−4)/(2−(√(x^2 +4)))) =lim_(x→0) ((−4)/(2−(√(x^2 +4)))) 2−(√(x^2 +4))=2−(4+x^2 )^(1/2) =2−2(1+(x^2 /4))^(1/2) =2−2(1+(1/2)(x^2 /4)+...)=−((x^2 /8)+all +ve yerms) hence limits exists and is infinity.
$$\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:<\mathrm{0}\:\mathrm{for}\:\mathrm{all}\:{x} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}=−\mathrm{4}\:\mathrm{near}\:{x}=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{4}}{\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}=\mathrm{2}−\left(\mathrm{4}+{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} =\mathrm{2}−\mathrm{2}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$=\mathrm{2}−\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+…\right)=−\left(\frac{{x}^{\mathrm{2}} }{\mathrm{8}}+{all}\:+{ve}\:\mathrm{yerms}\right) \\ $$$$\mathrm{hence}\:\mathrm{limits}\:\mathrm{exists}\:\mathrm{and}\:\mathrm{is}\:\mathrm{infinity}. \\ $$
Commented by prakash jain last updated on 13/Jan/18
both LHL and RHL tend to +infinity

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