# Thread: Help in finding the Limits..

1. ## Help in finding the Limits..

I need help determining the limits of the following problems.
Determine the following limits(if the limit exists otherwise explain why it does not) without using the L'Hospital's rule...

1) Lim [sqrt(4+x) - 2)] / [x]
x->0

2) Lim [ x^2 - 1] / [|x-1|]
x->1

3) Lim [sqrt(x) . cos (1/x) ] ......using L'Hospital's rule...
x->0

2. Originally Posted by Vedicmaths
2) Lim [ x^2 - 1] / [|x-1|]
x->1
$\lim_{x \to 1} \frac{x^2 - 1}{|x - 1|}$

As x approaches 1 from the left:
$\lim_{x \to 1^-} \frac{x^2 - 1}{-(x - 1)} = \lim_{x \to 1^-} -(x + 1) = -2$

As x approaches 1 from the right:
$\lim_{x \to 1^+} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^+} (x + 1) = 2$

Since these two are not the same, the limit does not exist.

-Dan

3. Originally Posted by topsquark

$= \lim_{x \to 0} \left ( \sqrt{\frac{4}{x^2} + \frac{1}{x}} - \frac{2}{x} \right )$

It should be clear at this point that the limit is 0.
At this point?

4. Originally Posted by Vedicmaths
3) Lim [sqrt(x) . cos (1/x) ] ......using L'Hospital's rule...
x->0
I really don't see how to apply L'Hopital's rule here:
$\lim_{x \to 0} \sqrt{x}~cos \left ( \frac{1}{x} \right )$

Let $y = \frac{1}{x}$. Then your limit is
$\lim_{x \to 0} \sqrt{x}~cos \left ( \frac{1}{x} \right ) = \lim_{y \to \infty} \frac{cos(y)}{\sqrt{y}}$

This is not an indeterminate form: cos(y) never reaches a limit, but is always between -1 and 1. $\sqrt{y}$, on the other hand, is unbounded and positive. Thus
$\lim_{x \to 0} \sqrt{x}~cos \left ( \frac{1}{x} \right ) = \lim_{y \to \infty} \frac{cos(y)}{\sqrt{y}} = 0$

-Dan

5. Hello
Originally Posted by Vedicmaths
1) Lim [sqrt(4+x) - 2)] / [x]
x->0
Originally Posted by topsquark
It should be clear at this point that the limit is 0.
Let $f(x)=\sqrt{4+x}$

Notice that $\lim_{x\to 0}\frac{\sqrt{4+x}-2}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)=\frac{1}{2\sqrt{4}}=\frac{1}{4}$

6. Originally Posted by flyingsquirrel
Hello

Let $f(x)=\sqrt{4+x}$

Notice that $\lim_{x\to 0}\frac{\sqrt{4+x}-2}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)=\frac{1}{2\sqrt{4}}=\frac{1}{4}$
I was thinking the limit was at infinity for some reason. Ugh!

Thanks for the catch!

-Dan