# Thread: Calculating limit on a trig problem

1. ## Calculating limit on a trig problem

Hi,
Needing help on the following limits, its been a long time since I have done trig, and have forgotten allot of trig properties.. any help would be appreciated on workings.

$\displaystyle \lim_{x \rightarrow 0} \frac{sinx^4}{sin^2x^2}$

$\displaystyle \lim_{x \rightarrow \infty} xsin \frac{1}{x}$

2. Originally Posted by Robb
Hi,
Needing help on the following limits, its been a long time since I have done trig, and have forgotten allot of trig properties.. any help would be appreciated on workings.

$\displaystyle \lim_{x \rightarrow 0} \frac{sinx^4}{sin^2x^2}$

Note that $\displaystyle \frac {\sin x^4}{\sin^2 x^2} = \frac {\sin x^4}{\sin^2 x^2} \cdot \frac {x^4}{x^4} = \frac {\sin x^4}{x^4} \cdot \left( \frac {x^2}{\sin x^2} \right)^2$

now take the limit. you should be able to take it from here
$\displaystyle \lim_{x \rightarrow \infty} xsin \frac{1}{x}$

Let $\displaystyle t = \frac 1x$. then we have $\displaystyle \lim_{t \to 0} \frac {\sin t}t$

i suppose you can take it from here.

3. Hello,
Originally Posted by Robb
Hi,
Needing help on the following limits, its been a long time since I have done trig, and have forgotten allot of trig properties.. any help would be appreciated on workings.

$\displaystyle \lim_{x \rightarrow 0} \frac{sinx^4}{sin^2x^2}$

$\displaystyle \lim_{x \rightarrow \infty} xsin \frac{1}{x}$

The key formula for these is $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}=1$

the first one can be written :
$\displaystyle \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \frac{x^4}{\sin^2(x^2)}=\lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \left(\frac{x^2}{\sin(x^2)}\right)^2$

The second one can be solved by making the substitution t=1/x

edit : lol too late ><

4. Originally Posted by Moo
edit : lol too late ><
well, you gave the key formula

5. Ok, Well i think I have got it now for the first question...

Using $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}=1$

I then took
$\displaystyle \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \frac{x^4}{\sin^2(x^2)}=\lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \left(\frac{x^2}{\sin(x^2)}\right)^2$

Setting t = x^4
$\displaystyle lim_{t \to 0} \frac{\sin(t)}{t} \cdot lim_{x \to 0} \left(\frac{x^2}{\sin(x^2)}\right)^2$
is equal to

$\displaystyle \lim_{x \to 0} \ 1 \cdot$
$\displaystyle \left(\frac{x^2}{\sin(x^2)}\right)^2$

Setting t = x^2

$\displaystyle \lim_{t \to 0} \ 1 \cdot \left(\frac{t}{\sin(t)}\right)^2$

Taking the inverse:
$\displaystyle \lim_{t \to 0} \frac{1}{1} \cdot \left(\frac{\sin(t)}{t}\right)^2 = \lim_{t \to 0} \ 1 \cdot (1)^2 = 1$

6. Originally Posted by Robb
Ok, Well i think I have got it now for the first question...

Using $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}=1$

I then took
$\displaystyle \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \frac{x^4}{\sin^2(x^2)}=\lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \left(\frac{x^2}{\sin(x^2)}\right)^2$

Setting t = x^4
$\displaystyle lim_{t \to 0} \frac{\sin(t)}{t} \cdot lim_{x \to 0} \left(\frac{x^2}{\sin(x^2)}\right)^2$
is equal to

$\displaystyle \lim_{x \to 0} \ 1 \cdot$
$\displaystyle \left(\frac{x^2}{\sin(x^2)}\right)^2$

Setting t = x^2

$\displaystyle \lim_{t \to 0} \ 1 \cdot \left(\frac{t}{\sin(t)}\right)^2$

Taking the inverse:
$\displaystyle \lim_{t \to 0} \frac{1}{1} \cdot \left(\frac{\sin(t)}{t}\right)^2 = \lim_{t \to 0} \ 1 \cdot (1)^2 = 1$
yes. but you didn't have to drag it out like that, just make two substitutions at once (if necessary, if you just said the limit was 1, i doubt your professor would mind).

$\displaystyle \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \left(\frac{x^2}{\sin(x^2)}\right)^2 = \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \lim_{x \to 0} \left(\frac{x^2}{\sin(x^2)}\right)^2$

Now let $\displaystyle t = x^4$ and $\displaystyle r = x^2$, then we have

$\displaystyle \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \lim_{x \to 0} \left(\frac{x^2}{\sin(x^2)}\right)^2 = \lim_{t \to 0} \frac{\sin t}t \cdot \lim_{r \to 0} \left(\frac r{\sin r}\right)^2 = 1 \cdot 1^2 = 1$

(By the way, you can't just take inverses like that)