# 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.

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

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

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.

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

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

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

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

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.

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

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

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

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

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

I then took
$\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
$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

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

Setting t = x^2

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

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

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

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

I then took
$\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
$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

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

Setting t = x^2

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

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

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

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

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

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