# Thread: Which trig limit did they use?

1. ## Which trig identity did they use?

The book I'm using seriously skips soooo many steps - it's hard to figure out how they get from one step to another.

2. Hello, RedSwiss!

The book I'm using seriously skips soooo many steps.
It's hard to figure out how they get from one step to another.

. . $\displaystyle \lim_{x\to\infty}\frac{1-\cos\left(\frac{1}{x}\right)}{\frac{1}{x^2}} \;=\;\lim_{x\to\infty}\frac{-\frac{1}{x^2}\sin\left(\frac{1}{x}\right)}{-\frac{2}{x^3}}$

They used L'Hopital's Rule.

We have: . $\dfrac{1-\cos(x^{-1})}{x^{-2}}$

The derivative of the numerator is: . $-\bigg[-\sin\left(x^{-1}\right)\bigg]\bigg[-x^{-2}\bigg] \;=\;-\frac{1}{x^2}\sin\left(\frac{1}{x}\right)$

The derivative of the denominator is: . $-2x^{-3} \;=\;-\frac{2}{x^3}$

3. in this problem L'Hospital's rule has been used. that is how you get the second expression from the first. according to LHospital's rule lim(x->a) n(x)/d(x)=lim(x->a) n'(x)/d'(x)=............=lim(x->a) n''''''....(x)/d'''''.....(x). differentiation can be carried out untill the undefined form of the given expression is got rid of.