Evaluate $$\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}$$ I tried to apply l'Hospital rule in order to get the limit to be equal to $$\lim_{x\to0}\frac{e^x-1}{2(e^x-\frac{x^2}{2}-x-1)^{-\frac{1}{3}}(e^x-x-1)}$$ but the new denominator has an indeterminate form itself and by repeatedly applying l'Hospital rule, it doesn't seem to help... This is where I got stuck.
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1$\begingroup$ Better use Taylor series and get the answer instantly as $1/\sqrt[3]{6}$. L'Hospital's Rule should be used if differentiation of numerator /denominator is easy and leads to simpler expressions. $\endgroup$– Paramanand Singh ♦Commented Aug 31, 2017 at 9:46
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$\begingroup$ Thank you! I stated it like that because it is an exercise from the l'Hospital rule chapter in my book and I haven't yet learnt Taylor series. $\endgroup$– ShroudCommented Aug 31, 2017 at 9:53
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1$\begingroup$ I provide an answer based on L'Hospital's Rule. Hope it helps. $\endgroup$– Paramanand Singh ♦Commented Aug 31, 2017 at 10:32
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$\begingroup$ It's exactly what I was looking for, thank you so much! $\endgroup$– ShroudCommented Aug 31, 2017 at 11:17
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3 Answers
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Use L'Hospital's Rule once to show that $$\lim_{x\to 0}\frac{e^{x}-1-x}{x^{2}}=\frac{1}{2}\tag{1}$$ and using this limit and L'Hospital's Rule once more show that $$\lim_{x\to 0}\dfrac{e^{x}-1-x-\dfrac{x^{2}}{2}}{x^{3}}=\frac{1}{6}\tag{2}$$ Now divide the numerator and denominator of the original expression (whose limit is to be evaluated here) by $x^{2}$ (for denominator write $x^{2}=(x^{3})^{2/3}$) and take limits (and use limits $(1),(2)$) to get the answer as $$\dfrac{\dfrac{1}{2}}{3\left(\dfrac{1}{6}\right) ^{2/3}}=\frac{1}{\sqrt[3]{6}}$$ One must always use certain algebraic manipulation before the application of advanced techniques like L'Hospital's Rule or Taylor series unless the question is especially suited to these techniques. Your direct application of L'Hospital's Rule only leads to complicated expressions.
answered Aug 31, 2017 at 10:31
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You can use $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5)$ $$\qquad{\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}=\\ \lim_{x\to0} \frac{(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))-x-1}{3((1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}=\\ \lim_{x\to0} \frac{(\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))}{3((\frac{x^3}{6}+\frac{x^4}{24}+o(x^5)))^{\frac{2}{3}}}=\\ \lim_{x\to0} \frac{x^2(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3(x^3(\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\ \lim_{x\to0} \frac{x^2(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3x^{3\times\frac{2}{3}}((\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\ \lim_{x\to0} \frac{(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3((\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\ \frac{\frac{1}{2}}{3(\frac{1}{6})^{\frac{2}{3}}}} $$
answered Aug 31, 2017 at 9:52
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$\begingroup$ You have a typo. You should have $24=4!$ instead of $120=5!$. Moreover you don't need to go beyond $o(x^{2})$ in numerator and beyond $o(x^{3})$ in denominator. $\endgroup$– Paramanand Singh ♦Commented Aug 31, 2017 at 10:39
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$\begingroup$ I fixed another minor typo in third last line. $\endgroup$– Paramanand Singh ♦Commented Aug 31, 2017 at 10:49
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$\begingroup$ @ParamanandSingh:thank you very much $\endgroup$ Commented Aug 31, 2017 at 10:50
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Using L'Hospital's rule. Note the limit can be transformed for convenience as follows: $$L=\frac13 \sqrt[3]{\lim_{x\to 0} \frac{(e^x-x-1)^3}{(e^x-\frac{x^2}{2}-x-1)^2}}=(L'H)=$$ $$\frac13 \sqrt[3]{\lim_{x\to 0} \frac{3(e^x-x-1)^2(e^x-1)}{2(e^x-\frac{x^2}{2}-x-1)(e^x-x-1)}}=(L'H)=$$ $$\frac{1}{\sqrt[3]{18}}\cdot \sqrt[3]{\lim_{x\to 0} \frac{2e^{2x}-3e^x-xe^x+1}{e^x-x-1}}=(L'H)=$$ $$\frac{1}{\sqrt[3]{18}}\cdot \sqrt[3]{\lim_{x\to 0} \frac{4e^{2x}-4e^x-xe^x}{e^x-1}}=(L'H)=$$ $$\frac{1}{\sqrt[3]{18}}\sqrt[3]{\lim_{x\to 0} \frac{8e^{2x}-5e^x-xe^x}{e^x}}=\frac{1}{\sqrt[3]{18}} \cdot \sqrt[3]{3}=\frac{1}{\sqrt[3]{6}}.$$
answered Aug 31, 2017 at 12:00