# Finding the limit

• Oct 10th 2011, 05:06 PM
bhuang
Finding the limit
Hi, can anyone help me approach the following problems:

Find the limit, or explain why it does not exist.

1) lim (1-|x|) / (x-1) as x-->1
2) lim ((sq. rt(1-h))-1) / h as h-->0

Using a computer or calculator, estimate the following limits. Use the technology to sketch the graph of the function to check the accuracy of your estimate.

lim (e^(2x) -1 ) / x as x-->0

I know for this last one it says to use a calculator or computer.. but then I really find no point in even doing the problem at all if we can just use a calculator. So is there a way to solve this limit algebraically? Is there an easy way to graph this as well? Or do I just have to sub in values and graph from the table of values?

Thanks!
• Oct 10th 2011, 05:18 PM
TheChaz
Re: Finding the limit
For the third one, you'll have more tools for computing this limit later (L'Hospital's rule)...
• Oct 10th 2011, 06:04 PM
Prove It
Re: Finding the limit
Quote:

Originally Posted by bhuang
Hi, can anyone help me approach the following problems:

Find the limit, or explain why it does not exist.

1) lim (1-|x|) / (x-1) as x-->1
2) lim ((sq. rt(1-h))-1) / h as h-->0

Using a computer or calculator, estimate the following limits. Use the technology to sketch the graph of the function to check the accuracy of your estimate.

lim (e^(2x) -1 ) / x as x-->0

I know for this last one it says to use a calculator or computer.. but then I really find no point in even doing the problem at all if we can just use a calculator. So is there a way to solve this limit algebraically? Is there an easy way to graph this as well? Or do I just have to sub in values and graph from the table of values?

Thanks!

For the first, note that $\displaystyle |x| = x \textrm{ if }x \geq 0$, so since you are making the limit approach 1, $\displaystyle 1 - |x| = 1 - x = -(x - 1)$.

Can you go from here?

For the second, try multiplying numerator and denominator by the numerator's conjugate...
• Oct 10th 2011, 06:07 PM
Deveno
Re: Finding the limit
with regards to question number 3, sort of:

if you are willing to accept that:

$e^x = 1 + x + x^2/2 + x^3/6 + ....+ x^n/n! +....$

(which you probably have not demonstrated yet, but it is true) then,

$e^{2x} = 1 + 2x + 4x^2/2 + 8x^3/6 +....+ 2^nx^n/n! +....$

and $e^{2x} - 1 = 2x + 4x^2/2 + 8x^3/6 +....+ 2^nx^n/n! +....$

so $\frac{e^{2x}-1}{x} = 2 + x(2x + 4x^3/3 +....+2^nx^{n-1}/n!+....)$

if, further, you are willing to accept that the stuff in the parentheses is finite (even though it has an infinite number of terms) then:

$\lim_{x \to 0} \frac{e^{2x}-1}{x} = 2$

now, that's a lot of stuff to take on faith, and so its appropriate to just make a guess based on how it behaves. my suggestion is that you guess "2", and remember this limit, so you can check later, when you have more theorems under your belt.
• Oct 11th 2011, 07:50 AM
HallsofIvy
Re: Finding the limit
For (2), $\lim_{h\to 0} \frac{\sqrt{1- h}- 1}{h}$, multiply both numerator and denominator by $\sqrt{1- h}+ 1$.