Check my limits Solution

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October 6th 2009, 12:13 PM
mj.alawami

Check my limits Solution

1) $lim (x\rightarrow 1) \frac{\sqrt{x}-1}{x-1}$
$\frac{1}{\sqrt{x}+1} =\frac{1}{2}$

2) $lim(x\rightarrow -1) \frac{\sqrt{h+5}-2}{h+1}$
$]\frac{1}{\sqrt{h+5}+2}=\frac{1}{4}$

3) $lim(x\rightarrow 1)\frac{^3\sqrt{x}-1}{x-1}$

$\frac{1}{^3\sqrt{x}+1}=\frac{1}{2}$

4) $lim(x\rightarrow 0) \frac{^3\sqrt{h+1}-1}{h}$

$\frac{1}{^3\sqrt{h+1}+1}=\frac{1}{2}$

If there are any mistake please correct them?

Thank you..
October 6th 2009, 12:44 PM
Grandad

Hello mj.alawami

Quote:

Originally Posted by mj.alawami

1) $lim (x\rightarrow 1) \frac{\sqrt{x}-1}{x-1}$
$\frac{1}{\sqrt{x}+1} =\frac{1}{2}$

2) $lim(x\rightarrow -1) \frac{\sqrt{h+5}-2}{h+1}$
$]\frac{1}{\sqrt{h+5}+2}=\frac{1}{4}$

3) $lim(x\rightarrow 1)\frac{^3\sqrt{x}-1}{x-1}$

$\frac{1}{^3\sqrt{x}+1}=\frac{1}{2}$

4) $lim(x\rightarrow 0) \frac{^3\sqrt{h+1}-1}{h}$

$\frac{1}{^3\sqrt{h+1}+1}=\frac{1}{2}$

If there are any mistake please correct them?

Thank you..

The first two are fine, but there's a problem with #3 and #4. In these you need to use

$x^3-1 = (x-1)(x^2+x+1)$

So #3 simplifies to $\frac{1}{x^{\frac23}+x^{\frac13}+1}$

... and #4 is similar.

Grandad
October 7th 2009, 01:48 AM
mj.alawami

Quote:

Originally Posted by Grandad

Hello mj.alawamiThe first two are fine, but there's a problem with #3 and #4. In these you need to use

$x^3-1 = (x-1)(x^2+x+1)$

So #3 simplifies to $\frac{1}{x^{\frac23}+x^{\frac13}+1}$

... and #4 is similar.

Grandad

Can you please show me the complete steps of rationalizing the numerator of #3 ?
October 7th 2009, 01:58 AM
mr fantastic

Quote:

Originally Posted by mj.alawami

Can you please show me the complete steps of rationalizing the numerator of #3 ?

My advice for #3 is to make the substitution $x = t^3$ so that you have $\lim_{t \rightarrow 1} \frac{t - 1}{t^3 - 1}$ . Now factorise the denominator, cancel the factor common to numerator and denominator and then take the limit.

I advise #4 be done in a similar way.
October 7th 2009, 07:58 AM
Grandad

Hello mj.alawami

Use Mr F's substitution, or you can do it directly, by writing $\sqrt[3]x$ as $x^{\frac13}$ .

Note first what happens when you multiply out the brackets from $(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)$ , using the laws of indices:

$(x^{\frac13}-1)\color{red}(x^{\frac23}+x^{\frac13}+1)\color{bla ck}= x^{\frac13}\color{red}(x^{\frac23}+x^{\frac13}+1)\ color{black}-1\color{red}(x^{\frac23}+x^{\frac13}+1)\color{blac k}=x + x^{\frac23}+x^{\frac13}-x^{\frac23}-x^{\frac13}-1=x-1$

So we can factorise the denominator: $x-1 =(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)$

So $\lim_{x\to 1}\frac{\sqrt[3]x-1}{x-1}=\lim_{x\to 1}\frac{x^{\frac13}-1}{(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)}$

$=\lim_{x\to 1}\frac{1}{x^{\frac23}+x^{\frac13}+1}$

$=\frac13$

Question 4 can be done in the same way. Start with

$\lim_{h\to0}\frac{\sqrt[3]{h+1}-1}{h}$

$=\lim_{h\to0}\frac{(h+1)^{\frac13}-1}{(h+1)-1}$

...which is basically the same as #3, with $x$ replaced by $(h+1)$ .

Grandad