susus
Dec 21, 2010, 04:25 AM
lim (tanx - sinx / x^3 ) = ?
x->0
I tried to do lopital , but, I did was stuck somewhere... I got 1/4.. the
galactus
Dec 21, 2010, 06:41 AM
Is that \lim_{x\to 0}\frac{tan(x)-sin(x)}{x^{3}} or what you have typed:
\lim_{x\to 0}[tan(x)-\frac{sin(x)}{x^{3}}]
I assume the former. Grouping symbols are important.
You can use L'Hopital, but it requires several applications.
First application:
\frac{-cos(x)+2tan(x)sec^{2}(x)}{3x^{2}}
second application:
\frac{sin(x)+2tan(x)sec^{2}(x)}{6x}
Third application:
\frac{cos(x)+2sec^{4}(x)+4tan^{2}(x)sec^{2}(x)}{6}
Now, letting x=0 in the numerator, we get:
\frac{3}{6}=\frac{1}{2}
Here is a non-L'Hopital way to go about it. I like this better.
Rewrite as:
\lim_{x\to 0}\frac{tan(x)(1-cos(x))}{x^{3}}
=\lim_{x\to 0}\frac{tan(x)}{x}\cdot \frac{1-cos(x)}{x^{2}}
Now, take each limit separately.
\lim_{x\to 0}\frac{tan(x)}{x}=\lim_{x\to 0}\left(\frac{sin(x)}{x}\right)\left(\frac{1}{cos( x)}\right)=1\cdot 1=1
Now for:
\lim_{x\to 0}\frac{1-cos(x)}{x^{2}}
=\lim_{x\to 0}\frac{1-cos(x)}{x^{2}}\cdot \frac{1+cos(x)}{1+cos(x)}
=\lim_{x\to 0}\frac{1-cos^{2}(x)}{x^{2}(1+cos(x))}
=\lim_{x\to 0}\frac{sin^{2}(x)}{x^{2}}\cdot \frac{1}{1+cos(x)}
=(1)(\frac{1}{2})=\frac{1}{2}
From the first limit we have 1. So, we get (1)(1)(\frac{1}{2})
Thus, the limit is 1/2
Note that \lim_{x\to 0}\frac{sin(x)}{x}=1 is a famous limit used often when proving other limits. It is not necessary to prove it here, but use it as a lemma, so to speak. It's proof usually involves the Squeeze Theorem.
As for \lim_{x\to 0}\frac{1}{1+cos(x)}, we can use an approximation. Knowing that cos(x)\approx 1-\frac{x^{2}}{2} for x near 0, we can sub this in and get:
\lim_{x\to 0}\frac{2}{4-x^{2}}.
Now, as can be seen, as x\to 0, we are left with 1/2.
Actually, we could have used this approximation, as well as sin(x)\approx x for x near 0, to prove from the outset.
Doing so, leads to
\lim_{x\to 0}\frac{tan(x)-sin(x)}{x^{3}}=\lim_{x\to 0}\frac{\frac{sin(x)}{cos(x)}-sin(x)}{x^{3}}
=\lim_{x\to 0}\frac{\frac{x}{1-\frac{x^{2}}{2}}-x}{x^{3}}=\lim_{x\to 0}\frac{1}{2-x^{2}}
Now, as x\to 0, we are left with... again... 1/2.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
When dealing with trig limits with x\to 0, then the following approximations will prove useful.
1-\frac{x^{2}}{2}\approx cos(x)
x\approx sin(x)
\frac{2-x^{2}}{2x}\approx tan(x)
\frac{2x}{2-x^{2}}\approx cot(x)
\frac{2}{2-x^{2}}\approx sec(x)
\frac{1}{x}\approx csc(x)
These are known as 'asymptotic' approximations.