Please help me solve lim (sin 2x)/x
x approaching zero.
I'm supposed to determine the limit graphically and then confirm algebraically.
Thank you.
Here's the graph:
Can you determine $\displaystyle \lim_{x\to 0}\frac{\sin(2x)}{x}$ now?
Here's a hint on how to verify it algebraically.
$\displaystyle \lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{2\sin(2x)}{2x}=2\lim_{x\to 0}\frac{\sin(2x)}{2x}$
This limit should look pretty familiar...
Does this help?
--Chris
Thank you very much for responding Chris. I only get part of it though. could you please tell me how to determine the limit just by looking at the graph? and also how do I solve the rest of the equation. I am very new to calculus so I would really appreciate your help.
To find the limit graphically, look at the value of the graph at $\displaystyle x=0$. In this case it appears to be 2, although it doesn't exist at 0 [otherwise we would be dividing by zero, which is illegal].
To finish it off, this is what we would do algebraically:
$\displaystyle \lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{2\sin(2x)}{2x}=2\lim_{x\to 0}\frac{\sin(2x)}{2x}$
This limit has the form of $\displaystyle \lim_{u\to 0}\frac{\sin(u)}{u}=1$
So let $\displaystyle u=2x$. Thus, as $\displaystyle x\to 0$, $\displaystyle u\to 0$ as well.
So our limit transforms from $\displaystyle 2\lim_{x\to 0}\frac{\sin(2x)}{2x}$ to $\displaystyle 2\lim_{u\to 0}\frac{\sin u}{u}=2(1)=\color{red}\boxed{2}$
Does this make sense?
--Chris
yes thank you that helped a lot.
could u help me with this other one please?
f(x)=(e^-x)/(x)
I'm supposed to
a: use graphs and tables to find lim x approaching infinity f(x)
b: lim x approaching negative infinity f(x)
c: Identify all horizontal asymptotes