Answer by user65203 for Calculate the following limit without L'Hopital
There is a simple way to address the denominator:$$(x+\tan x-\sin2x)'=1+\tan^2x+1-2\cos 2x=\tan^2x+4\sin^2x=\sin^2x\left(\frac1{\cos^2x}+4\right).$$The second factor will tend to $5$ and the first can...
View ArticleAnswer by trancelocation for Calculate the following limit without L'Hopital
$$\begin{eqnarray*} \frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)} & = & \underbrace{\frac{1}{e^x+\sqrt{1+2x+2x^2}}}_{\mbox{harmless}}\cdot \frac{e^{2x}-(1+2x+2x^2)}{x+\tan (x)-\sin (2x)}...
View ArticleAnswer by user for Calculate the following limit without L'Hopital
HINTBy Taylor's series we have that$$\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}=\frac{\overbrace{1+x+\frac12 x^2+\frac16x^3+o(x^3)}^{\color{red}{e^x}}-(\overbrace{1+x+\frac12x^2-\frac12...
View ArticleAnswer by user65203 for Calculate the following limit without L'Hopital
It is not so "very awful" as you say, if you tame the computation.For the second term of the numerator, you can work this out...
View ArticleCalculate the following limit without L'Hopital
$\lim\limits_{x\to 0}\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}$ I know how to count this limit with the help of l'Hopital rule. But it is very awful, because I need 3 times derivate it. So,...
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