Limit problem not making sense

[Specter]

Homework Statement

$\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}$

Homework Equations

The Attempt at a Solution

I know the limit as x approaches 0 isn't supposed to be a fraction but I can't get the x approaches 0 under the lim.

I couldn't get some of this typed out in latex, it just wouldn't work for me. But here's the rest of the work. The square root is only over the x+1's.


$\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}$

$\frac {lim} {x→0} \frac {(√x+1 -1)(√x+1 +1)} {x(√x+1 +1)}$

$\frac {lim} {x→0} \frac {x} {x(√x+1 +1)}$

$=\frac {1} {√1+1}$

=0.5

I have a few questions. From step 2 to step 3 the numerator gets reduced only to x. I don't know how.

When multiplying the numerator by it's conjugate, does it always result in the numerator equaling 1?

For step 4, does x just get factored out of the denominator and the numerator of x just gets changed to 1?

 

[fresh_42]

Homework Statement

$\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}$

Homework Equations

The Attempt at a Solution

I know the limit as x approaches 0 isn't supposed to be a fraction but I can't get the x approaches 0 under the lim.

I couldn't get some of this typed out in latex, it just wouldn't work for me. But here's the rest of the work. The square root is only over the x+1's.$\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}$

$\frac {lim} {x→0} \frac {(√x+1 -1)(√x+1 +1)} {x(√x+1 +1)}$

$\frac {lim} {x→0} \frac {x} {x(√x+1 +1)}$

$=\frac {1} {√1+1}$

=0.5

I have a few questions. From step 2 to step 3 the numerator gets reduced only to x. I don't know how.

Do you know the formula $(a-b)\cdot (a+b) =a^2-b^2\,?$ Apply it to $a=\sqrt{x+1}$ and $b=1$.

When multiplying the numerator by it's conjugate, does it always result in the numerator equaling 1?

a.) No. b) What do you mean by $1$? It resulted in an $x$ in this case. c) Try different values of $a,b$ in the formula above.

For step 4, does x just get factored out of the denominator and the numerator...

Yes.

... of or x just gets changed to 1?

No.

 

[Delta2]

For step 4 its like having $\frac{x}{xf(x)}$ isn't that equal to $\frac{1}{f(x)}$? The x's get simplified from the numerator and the denominator and leave just 1 behind, its like having $\frac{x}{x}$ which is 1.

So it is not x changed to 1, it is just $\frac{x}{x}$ which gets changed to its equal which is 1.

 

[SammyS]

Homework Statement

$\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}$I know the limit as x approaches 0 isn't supposed to be a fraction but I can't get the x approaches 0 under the lim.

I couldn't get some of this typed out in latex, it just wouldn't work for me. But here's the rest of the work. The square root is only over the x+1's.

As regards the LaTeX ...

To get the limit to display correctly use:
$\lim_{x \to 0} \frac {\sqrt {x+1}-1} {x}$ which gives $\lim_{x \to 0} \frac {\sqrt {x+1}-1} {x}$

OR use \displaystyle to improve readability.

$\displaystyle \lim_{x \to 0} \frac {\sqrt {x+1}-1} {x}$ which gives $\displaystyle \lim_{x \to 0} \frac {\sqrt {x+1}-1} {x}$

 

[mjc123]

If you aren't happy with the result, check by doing it another way. Try expanding (1+x)^1/2 as a Taylor series.

 

[SammyS]

@Specter :
Rather than simply marking the thread as being solved, please respond to the replies to the thread.

It's nice that you "Like" the responses of @Delta2 and @fresh_42 , but replies showing that you actually understand their posts would be a much better demonstration of showing that you appreciate the effort they put forth to respond to your questions.

 

[Master1022]

Homework Statement

$\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}$
I know the limit as x approaches 0 isn't supposed to be a fraction but I can't get the x approaches 0 under the lim.

I couldn't get some of this typed out in latex, it just wouldn't work for me. But here's the rest of the work. The square root is only over the x+1's.

Another method you may want to try is L'Hopital's Rule which is an efficient method of calculating some of these limits questions when you have fractions...

l'hospital's Rule states that
$$ \displaystyle \lim_{x \to c} \frac {f(x)} {g(x)} = \displaystyle \lim_{x \to c} \frac {f^n(x)} {g^n(x)} $$

where n is the nth derivative...

so you could differentiate the numerator and denominator independently and form the new fraction. Then you can plug zero into get the answer. Sometimes you may have to differentiate again as well.

Hope that is of some help. Just wanted to provide another way of tackling these problems.

 

[Specter]

Do you know the formula $(a-b)\cdot (a+b) =a^2-b^2\,?$ Apply it to $a=\sqrt{x+1}$ and $b=1$.

a.) No. b) What do you mean by $1$? It resulted in an $x$ in this case. c) Try different values of $a,b$ in the formula above.

Yes.No.

Thank you, this helped me make some sense of it.