# Evaluating limits

• Jan 19th 2010, 11:51 PM
Makall
Evaluating limits
lim x->0 (x cos x - sin x) / (x - sin x)
lim x->0 (-x sin x) / (1 - cos x)
lim x->0 (- sin x - x cos x) / (sin x)
lim x->0 (-2cos x + x sin x) / (cos x)
= -2

This is what's written in my book (Numerical Analysis). It has a few review questions on limits but it doesn't go in to detail so I'm not sure how to go from step 1 to step 2.
• Jan 20th 2010, 12:08 AM
Prove It
Quote:

Originally Posted by Makall
lim x->0 (x cos x - sin x) / (x - sin x)
lim x->0 (-x sin x) / (1 - cos x)
lim x->0 (- sin x - x cos x) / (sin x)
lim x->0 (-2cos x + x sin x) / (cos x)
= -2

This is what's written in my book (Numerical Analysis). It has a few review questions on limits but it doesn't go in to detail so I'm not sure how to go from step 1 to step 2.

1. Since this tends to $\frac{0}{0}$, you can use L'Hospital's Rule.

$\lim_{x \to 0} \frac{x\cos{x} - \sin{x}}{x - \sin{x}} = \lim_{x \to 0}\frac{\frac{d}{dx}(x\cos{x} - \sin{x})}{\frac{d}{dx}(x - \sin{x})}$

$= \lim_{x \to 0}\frac{\cos{x} - x\sin{x} - \cos{x}}{1 - \cos{x}}$

$= \lim_{x \to 0}\frac{-x\sin{x}}{1 - \cos{x}}$

This still tends to $\frac{0}{0}$, so use L'Hospital's Rule again.

$= \lim_{x \to 0}\frac{\frac{d}{dx}(-x\sin{x})}{\frac{d}{dx}(1 - \cos{x})}$

$= \lim_{x \to 0}\frac{-\sin{x} - x\cos{x}}{\sin{x}}$

This still tends to $\frac{0}{0}$, so use L'Hospital's Rule again.

$= \lim_{x \to 0}\frac{\frac{d}{dx}(-\sin{x} - x\cos{x})}{\frac{d}{dx}(\sin{x})}$

$= \lim_{x \to 0}\frac{-\cos{x} - \cos{x} + x\sin{x}}{\cos{x}}$

$= \lim_{x \to 0}\frac{x\sin{x} - 2\cos{x}}{\cos{x}}$

$= \frac{-2}{1}$

$= -2$.
• Jan 20th 2010, 12:13 AM
Prove It
Quote:

Originally Posted by Makall
lim x->0 (x cos x - sin x) / (x - sin x)
lim x->0 (-x sin x) / (1 - cos x)
lim x->0 (- sin x - x cos x) / (sin x)
lim x->0 (-2cos x + x sin x) / (cos x)
= -2

This is what's written in my book (Numerical Analysis). It has a few review questions on limits but it doesn't go in to detail so I'm not sure how to go from step 1 to step 2.

2. $\lim_{x \to 0}\frac{-x\sin{x}}{1 - \cos{x}}$

Since this tends to $\frac{0}{0}$, you can use L'Hospital's Rule.

$= \lim_{x \to 0}\frac{\frac{d}{dx}(-x\sin{x})}{\frac{d}{dx}(1 - \cos{x})}$

$= \lim_{x \to 0}\frac{-\sin{x} - x\cos{x}}{\sin{x}}$

Since this tends to $\frac{0}{0}$, use L'Hospital's Rule again.

$= \lim_{x \to 0}\frac{\frac{d}{dx}(-\sin{x} - x\cos{x})}{\frac{d}{dx}(\sin{x})}$

$= \lim_{x \to 0}\frac{-\cos{x} - \cos{x} + x\sin{x}}{\cos{x}}$

$= \lim_{x \to 0}\frac{-2\cos{x} + x\sin{x}}{\cos{x}}$

$= \frac{-2}{1}$

$= -2$.

BTW I thought the second step was a whole new question - could someone delete this post please?