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I want to calculate the limit which is above without using L'hopital's rule ;

$$\lim_{x\rightarrow0} \frac{e^x-1}{\sin(2x)}$$

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    Multiply and divide by $x$ and write it as the quotient of two known limits. – 2017-02-01
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    Why do you want to do this without L'Hopital rule? This makes no sense. – 2017-02-01
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    thanks for the answers, i got it :) – 2017-02-01
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    @Wolfram a lot of teachers prohibit students to use anything he/she hasn't teached yet, even if the student is doing the course for the second time or studied at home. This is pure nonsense, but happens. – 2017-02-02

3 Answers 3

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Using the fact that $$\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } =1 } \\ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } =1 } $$ we can conclude that $$\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ \sin { 2x } } } =\frac { 1 }{ 2 } \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } \frac { 2x }{ \sin { 2x } } } =\frac { 1 }{ 2 } $$

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    I would put $\frac{1}{2}$ on the other side of $\lim\limits_{x\to0}$, but it's just nitpicking. – 2017-02-01
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As an alternative and admittedly more mechanical approach, you can series expand top and bottom:

$$\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ \sin2x }} =\lim _{ x\rightarrow 0 }\frac{1+x+O(x^2)-1}{2x+O(x^3)}=\frac12$$

This will work for most fractions if the variable is tending to $0$. In fact, l'Hopital's rule essentially comes from series expanding $\frac{f(x)}{g(x)}$, so in a sense we are doing the same thing.

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Equivalents: $\;\mathrm e^x-1\sim_0 x$, $\;\sin 2x\sim_0 2x$, so $\;\dfrac{\mathrm e^x-1}{\sin 2x}\sim_0\dfrac{x}{2x}=\dfrac12.$

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    Of course, this is just haqnatural's answer written up differently. – 2017-02-01
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    Of course. But knowing the rules of asymptotic analysis makes it often much shorter, without having to fight with irrelevant details. That's what I wanted to point out. – 2017-02-01