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Question Number 146119 by liberty last updated on 11/Jul/21

          determinant (((lim_(x→0)  ((√(5x^2 +4x^4 ))/(3x)) =?)),((lim_(x→0)  (x^3 /( (√(x^6 +3x^7 )))) =?)))

$$\:\:\:\:\:\:\:\:\:\begin{array}{|c|c|}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{4}} }}{\mathrm{3}{x}}\:=?}\\{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} }{\:\sqrt{{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{7}} }}\:=?}\\\hline\end{array} \\ $$

Commented by ajfour last updated on 11/Jul/21

you have been on this forum,  quite a while, haven′t you  grasped this much- i mean  order of infinitesimals..

$${you}\:{have}\:{been}\:{on}\:{this}\:{forum}, \\ $$$${quite}\:{a}\:{while},\:{haven}'{t}\:{you} \\ $$$${grasped}\:{this}\:{much}-\:{i}\:{mean} \\ $$$${order}\:{of}\:{infinitesimals}.. \\ $$

Commented by EDWIN88 last updated on 11/Jul/21

  .does everyone who posts on this forum say he doesn't understand?  you are very wrong.  many who post just so there is material for discussion.

$$ \\ $$.does everyone who posts on this forum say he doesn't understand? you are very wrong. many who post just so there is material for discussion.

Answered by gsk2684 last updated on 11/Jul/21

lim_(x→0^− ) ((∣x∣(√(5+4x^2 )))/(3x))=lim_(x→0) ((−x(√(5+4x^2 )))/(3x))=((−(√5))/3)  lim_(x→0^+ ) ((∣x∣(√(5+4x^2 )))/(3x))=lim_(x→0) ((x(√(5+4x^2 )))/(3x))=((√5)/3)  limit does not exist   lim_(x→0^− ) (x^3 /(∣x^3 ∣(√(1+3x))))=lim_(x→0) (x^3 /(−x^3 (√(1+3x))))=−1   lim_(x→0^+ ) (x^3 /(∣x^3 ∣(√(1+3x))))=lim_(x→0) (x^3 /(x^3 (√(1+3x))))=1  limit does not exist

$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{\mid{x}\mid\sqrt{\mathrm{5}+\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{3}{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}\sqrt{\mathrm{5}+\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{3}{x}}=\frac{−\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mid{x}\mid\sqrt{\mathrm{5}+\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{3}{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{5}+\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{3}{x}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$${limit}\:{does}\:{not}\:{exist}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{{x}^{\mathrm{3}} }{\mid{x}^{\mathrm{3}} \mid\sqrt{\mathrm{1}+\mathrm{3}{x}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{3}} }{−{x}^{\mathrm{3}} \sqrt{\mathrm{1}+\mathrm{3}{x}}}=−\mathrm{1}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{x}^{\mathrm{3}} }{\mid{x}^{\mathrm{3}} \mid\sqrt{\mathrm{1}+\mathrm{3}{x}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} \sqrt{\mathrm{1}+\mathrm{3}{x}}}=\mathrm{1} \\ $$$${limit}\:{does}\:{not}\:{exist}\:\: \\ $$

Answered by ajfour last updated on 11/Jul/21

(i)   L=lim_(x→0) ((√(5+4x^2 ))/(3s))=((√5)/3)s          s=1  if x>0 ,          s=−1  if x<0  so limit dont exist at x=0.  (ii) L=lim_(x→0) (s/( (√(1+3x))))=s   ⇒ limit dont exist.

$$\left({i}\right)\:\:\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{5}+\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{3}{s}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}{s} \\ $$$$\:\:\:\:\:\:\:\:{s}=\mathrm{1}\:\:{if}\:{x}>\mathrm{0}\:, \\ $$$$\:\:\:\:\:\:\:\:{s}=−\mathrm{1}\:\:{if}\:{x}<\mathrm{0} \\ $$$${so}\:{limit}\:{dont}\:{exist}\:{at}\:{x}=\mathrm{0}. \\ $$$$\left({ii}\right)\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{s}}{\:\sqrt{\mathrm{1}+\mathrm{3}{x}}}={s} \\ $$$$\:\Rightarrow\:{limit}\:{dont}\:{exist}. \\ $$

Answered by hknkrc46 last updated on 11/Jul/21

(1) lim_(x→0)  ((√(5x^2  + 4x^4 ))/(3x)) = lim_(x→0)  ((√(x^2 (5 + 4x^2 )))/(3x))  = lim_(x→0)  ((x(√(5 + 4x^2 )))/(3x)) = lim_(x→0)  ((√(5 + 4x^2 ))/3) = ((√5)/3)  (2) lim_(x→0)  (x^3 /( (√(x^6  + 3x^7 )))) = lim_(x→0)  (x^3 /( (√(x^6 (1 + 3x)))))  = lim_(x→0)  (x^3 /(x^3 (√(1 + 3x)))) = lim_(x→0)  (1/( (√(1 + 3x)))) = 1

$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}\boldsymbol{{x}}^{\mathrm{2}} \:+\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{4}} }}{\mathrm{3}\boldsymbol{{x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\boldsymbol{{x}}^{\mathrm{2}} \left(\mathrm{5}\:+\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \right)}}{\mathrm{3}\boldsymbol{{x}}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{{x}}\sqrt{\mathrm{5}\:+\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} }}{\mathrm{3}\boldsymbol{{x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}\:+\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} }}{\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\:\sqrt{\boldsymbol{{x}}^{\mathrm{6}} \:+\:\mathrm{3}\boldsymbol{{x}}^{\mathrm{7}} }}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\:\sqrt{\boldsymbol{{x}}^{\mathrm{6}} \left(\mathrm{1}\:+\:\mathrm{3}\boldsymbol{{x}}\right)}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\boldsymbol{{x}}^{\mathrm{3}} \sqrt{\mathrm{1}\:+\:\mathrm{3}\boldsymbol{{x}}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:\mathrm{3}\boldsymbol{{x}}}}\:=\:\mathrm{1} \\ $$

Commented by lyubita last updated on 11/Jul/21

wrong answers

$${wrong}\:{answers} \\ $$

Commented by henderson last updated on 14/Jul/21

0 ∉ Domain of  expressions .  so you need to compare lim_(x→0^− )      and  lim_(x→0^+ )    .

$$\mathrm{0}\:\notin\:\boldsymbol{\mathrm{Domain}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{expressions}}\:. \\ $$$$\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{compare}}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\:\:\:\:\boldsymbol{\mathrm{and}}\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\:\:. \\ $$

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