# How could I calculate this limit without using L'Hopital's Rule $\lim_{x\rightarrow0} \frac{e^x-1}{\sin(2x)}$?

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

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

• Multiply and divide by $x$ and write it as the quotient of two known limits.
– user228113
Commented Feb 1, 2017 at 20:01
• Why do you want to do this without L'Hopital rule? This makes no sense. Commented Feb 1, 2017 at 20:07
• thanks for the answers, i got it :) Commented Feb 1, 2017 at 20:15
• @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. Commented Feb 2, 2017 at 1:25

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 $$\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ \sin { 2x } } } =\frac { 1 }{ 2 } \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } \frac { 2x }{ \sin { 2x } } } =\frac { 1 }{ 2 }$$

• I would put $\frac{1}{2}$ on the other side of $\lim\limits_{x\to0}$, but it's just nitpicking. Commented Feb 1, 2017 at 22:52

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.

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

• Of course, this is just haqnatural's answer written up differently. Commented Feb 1, 2017 at 20:55
• 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. Commented Feb 1, 2017 at 21:49