# Thread: [SOLVED] Limits with trig functions

1. ## [SOLVED] Limits with trig functions

i really hate finding the limits of equations with [for ex.] sinx or xsin x...it throws me off and i dont know what to do...im going ovver some problems over some practice problems because my test is either tomorrow or thursday.

lim x->0 x cscx + 1 / x cscx

the only thing i would have to guess is thati might have to do something with 1/sinx. i may be wrong.

2. An important trig limit that you need to use in this problem is $\displaystyle \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$ See if you can apply that to your problem.

3. wait wait wait, i think i got it...since cscx = 1/sin x, you can [since it is multiplied by x again...] you have wht you have provided...and then that equals one...meaning 1+1/1 = 2, the limit!

Originally Posted by Jameson
An important trig limit that you need to use in this problem is $\displaystyle \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$ See if you can apply that to your problem.

4. Originally Posted by >_<SHY_GUY>_<
i really hate finding the limits of equations with [for ex.] sinx or xsin x...it throws me off and i dont know what to do...im going ovver some problems over some practice problems because my test is either tomorrow or thursday.

lim x->0 x cscx + 1 / x cscx

the only thing i would have to guess is thati might have to do something with 1/sinx. i may be wrong.

Is the expression $\displaystyle \frac{\text{cosec} (x) + 1}{x \, \text{cosec} x} ?$ If so, then it can be re-written as $\displaystyle \frac{1 + \sin x}{x}$ and the limit clearly does not exist.