# Limits

• October 30th 2008, 09:54 AM
geeko
Limits
hey I got this question:

Question
lim x sin1/x=0
x->0
____________

-1<=sin(1/x)<=1

-x<=xsin(1/x)<=x

lim -x = 0
x->0

lim x=0
x->0

x->0

__________________

I just want to know if my working and answer is correct!tnx
• October 30th 2008, 10:08 AM
Chris L T521
Quote:

Originally Posted by geeko
hey I got this question:

Question
lim x sin1/x=0
x->0
____________

-1<=sin(1/x)<=1

-x<=xsin(1/x)<=x

lim -x = 0
x->0

lim x=0
x->0

x->0

__________________

I just want to know if my working and answer is correct!tnx

The answer is correct, and the working seems to be correct too [since I'm not an expert on applying the Squeeze Theorem]

Here's another approach:

$\lim_{x\to0}x\sin\left(\frac{1}{x}\right)$

Let $z=\frac{1}{x}\implies x=\frac{1}{z}$

Now, as $x\to 0,~z\to\infty$

Thus, $\lim_{x\to 0}x\sin\left(\frac{1}{x}\right)=\lim_{z\to\infty}\ frac{\sin z}{z}$ which converges to zero.

--Chris
• October 30th 2008, 10:43 AM
geeko
Quote:

Originally Posted by Chris L T521
The answer is correct, and the working seems to be correct too [since I'm not an expert on applying the Squeeze Theorem]

Here's another approach:

$\lim_{x\to0}x\sin\left(\frac{1}{x}\right)$

Let $z=\frac{1}{x}\implies x=\frac{1}{z}$

Now, as $x\to 0,~z\to\infty$

Thus, $\lim_{x\to 0}x\sin\left(\frac{1}{x}\right)=\lim_{z\to\infty}\ frac{\sin z}{z}$ which converges to zero.

--Chris

Quote:

lim f(x) where |f(x)-1| <=x2(p.s x square) , x not equal to 0
x->0

multiply by x ..... -x <= x sin(1/x) <= x

lim-x=0
x->0

lim x=0
x->0

therefore lim x sin(1/x)=0
x->0

Is it correct too?