1. ## 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!

2. ## Re: Finding the limit

For the third one, you'll have more tools for computing this limit later (L'Hospital's rule)...

3. ## Re: Finding the limit

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...

4. ## 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.

5. ## Re: Finding the limit

For (2), $\lim_{h\to 0} \frac{\sqrt{1- h}- 1}{h}$, multiply both numerator and denominator by $\sqrt{1- h}+ 1$.