1. ## limits

how do you solve lim ( sin x/ x) as

(a) x tends to 1
(b)x tends to infinity?

without using l'hosipital rule..

can i say that for part (a) since the lim of top and bottom of the fraction exiist, lim ( sinx/ x) = lim ( sinx ) lim (1/x)? then im stuck alr...
thanks

2. Originally Posted by alexandrabel90
how do you solve lim ( sin x/ x) as

(a) x tends to 1
(b)x tends to infinity?

without using l'hosipital rule..

can i say that for part (a) since the lim of top and bottom of the fraction exiist, lim ( sinx/ x) = lim ( sinx ) lim (1/x)? then im stuck alr...
thanks

Yes, you can say that because the limits exists AND ARE FINITE, so $\lim_{x\to 1}\frac{\sin x}{x}=\sin 1$

Now for (b) prove the following easy lemma:

If $f(x)\xrightarrow [x\to x_0] {} 0\,\,and \,\,g(x)$ bounded in some neighborhood of $x_0$ , then $\lim_{x\to x_0}f(x)g(x)=0$ , and this lemma includes the case $x_0=\infty$

Tonio

Pss. BTW, the condition not to use L'H rule is pointless: you can NOT use L'H in none of both cases.

3. Could you use the Sandwich Theorem for part b?

$\frac{-1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}$.

Letting $x \rightarrow \infty$ gives you...

$0 \leq \frac{\sin(x)}{x} \leq 0$ so $\lim_{x \rightarrow \infty} \frac{\sin(x)}{x} = 0$..?

(This is sort of directed to Tonio by the way but is another method to look at)

4. Originally Posted by tonio
you can NOT use L'H in none of both cases.
Nice.

Could you use the Sandwich Theorem for part b?

$\frac{-1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}$.

Letting $x \rightarrow \infty$ gives you...

$0 \leq \frac{\sin(x)}{x} \leq 0$ so $\lim_{x \rightarrow \infty} \frac{\sin(x)}{x} = 0$..?

(This is sort of directed to Tonio by the way but is another method to look at)

Cool!...and correct, of course.

Tonio