#2**+15 **

lim x→ 1 [ (5 - x)^(1/2) - 2 ] / [(10 - x)^(1/2) -3 ]

Here, we have a 0/0 situation when we try to evaluate the limit. In this case, we can use something called "L'Hopital's Rule" to evaluate the limit.

Take the derivative of the numerator and the denominator.....this gives

lim x→ 1 (-1/2)(5-x)^(-1/2) / (-1/2)(10 - x)^(-1/2) =

lim x→ 1 (10-x)^(1/2)/(5-x)^(1/2) now, take the limit and we get

(10-1)^(1/2) / (5 -1)^(1/2) = 9^(1/2) / 4^(1/2) = 3 / 2

CPhill Oct 19, 2014

#1**+5 **

**I wish you would put your questions in the right way around 315. **

$$\\lim \limits_{x\rightarrow 1}\;\;\frac{\sqrt{5-x}-2}{\sqrt{10-x}-3}\\\\

=lim \limits_{x\rightarrow 1}\;\;\frac{(\sqrt{5-x}-2)(\sqrt{10-x}+3)}{(\sqrt{10-x}-3)(\sqrt{10-x}+3)}\\\\

=lim \limits_{x\rightarrow 1}\;\;\frac{(\sqrt{5-x}-2)(\sqrt{10-x}+3)}{1-x}\\\\$$

WELL THAT DIDN'T HELP AT ALL - SO DON'T DO THAT LOL

I CAN'T EVEN GET THE LATEX TO DISPLAY PROPERLY

I'VE DONE THIS ONE PROPERLY (WITHOUT L'HOPITAL'S RULE) IN A LATER POST ON THIS THREAD.

Melody Oct 19, 2014

#2**+15 **

Best Answer

lim x→ 1 [ (5 - x)^(1/2) - 2 ] / [(10 - x)^(1/2) -3 ]

Here, we have a 0/0 situation when we try to evaluate the limit. In this case, we can use something called "L'Hopital's Rule" to evaluate the limit.

Take the derivative of the numerator and the denominator.....this gives

lim x→ 1 (-1/2)(5-x)^(-1/2) / (-1/2)(10 - x)^(-1/2) =

lim x→ 1 (10-x)^(1/2)/(5-x)^(1/2) now, take the limit and we get

(10-1)^(1/2) / (5 -1)^(1/2) = 9^(1/2) / 4^(1/2) = 3 / 2

CPhill Oct 19, 2014

#3**+10 **

Thanks Chris, I keep forgetting about L'Hopital's Rule !!

https://www.math.hmc.edu/calculus/tutorials/lhopital/

there are also a couple of you tube clips on this.

That might be enough help 315. Let us know if it is not.

Melody Oct 19, 2014

#5**+5 **

the first one looks straight forward. It is 0/0 so you just use L'Hopital's Rule

The second and third one we cannot read.

The last one is really easy - i did it below.

Melody Oct 19, 2014

#6**+5 **

This is the last one - it is really easy

$$\displaystyle \lim_{x\rightarrow -1}\;\;\dfrac{|x+1|}{x^2+1}=

\dfrac{|-1+1|}{(-1)^2+1}=\dfrac{0}{1+1}=0$$

this is the graph if you are curious.

https://www.desmos.com/calculator/zhujqtjhpm

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

SORRY I HAVE ANSWERED A DIFFERENT QUESTION!

**YES - THE ANSWER YOU HAVE SUBMITTED BELOW IS CORRECT!**

THIS IS THE GRAPH

Melody Oct 19, 2014

#7**0 **

sorry for the pictuer .. I think this is better

I want the answers for these questions

and this is the last question you read it wrong ..

tell me if my answer is correct

xvxvxv Oct 19, 2014

#8**+5 **

I think that the 3rd one is

$$\displaystyle\lim_{x\rightarrow0} \;\frac{tan^{-1}x}{x}$$ Is this correct?

Anyway we have a situation of 0/0 so we can use L'Hopital's Rule

Let

$$\begin{array}{rll}

\alpha &=& tan^{-1}x\\

x &=& tan\alpha\\

\frac{dx}{d\alpha} &=& sec^2\alpha\\

\frac{d\alpha}{dx} &=& cos^2\alpha\\

\frac{d\alpha}{dx} &=& cos^2(tan^{-1}x)\\

\frac{d\alpha}{dx} &=& \frac{1}{1+x^2}\\

\end{array}$$

$$\displaystyle\lim_{x\rightarrow0} \;\frac{tan^{-1}x}{x}\\\\

=\displaystyle\lim_{x\rightarrow0} \;\frac{\left(\frac{1}{1+x^2}\right)}{1}\\\\

=\frac{1}{1+0^2}

=1$$

check

https://www.desmos.com/calculator/wxfrv3yegb

Yes that is correct.

Melody Oct 19, 2014

#9**+5 **

2nd one

$$\\\displaystyle\lim_{x\rightarrow1}\;\frac{x-1}{lnx}\\\\

$on first inspection this is 0/0 so you can use L'Hopital's Rule$\\\\\\

=\displaystyle\lim_{x\rightarrow1}\;\frac{1}{1/x}\\\\

=\frac{1}{1/1}\\\\

=1$$

Melody Oct 19, 2014

#10

#13**+5 **

I wrote that LOL

It is slang.

"Beats me" means "I do not know"

I was just attracting Alan's attention because I don't know the answer.

Melody Oct 19, 2014

#15**+10 **

Some can be done without L'Hopitals's rule. For example:

(x^{2} - 1)/(x^{2} - 2x +1) = (x - 1)(x + 1)/(x - 1)^{2} = (x + 1)/(x - 1) As x goes to 1 from above this goes to +∞; as x goes to 1 from below it goes to -∞.

When x is small then tan(x) → x, so tan^{-1}(x) → x (x in radians). Therefore tan^{-1}(x)/x → 1

.

Alan Oct 19, 2014

#16**+10 **

Another without L'Hopital's rule (though you need to know the expansion of ln(1-z)):

(x-1)/ln(x)

Let x = 1-z so (x-1)/ln(x) becomes -z/ln(1 - z)

As x → 1 , z → 0

Expand ln(1 - z) in an infinite series: ln(1 - z) = -z - z^{2}/2 - z^{3}/3 - z^{4}/4 - ...

so -z/ln(1 - z) = 1/(1 + z/2 + z^{2}/3 + z^{3}/4 + ...)

When z = 0 this becomes 1/(1 + 0 + 0 + ...) = 1

-z/ln(1 - z) tends to 1 as z tends to 0.

Hence (x-1)/ln(x) tends to 1 as x tends to 1

Edited to take the correct limit!

.

Alan Oct 19, 2014

#17**+13 **

highlight yellow question.

Yes I took a leap there. I shall show you. :)

Consider

$$cos^2(tan^{-1}x)\\\\$$

Remember that $$tan^{-1}x$$ is an angle

So I used this triangle

$$cos^2(tan^{-1}x)=\left(\frac{1}{\sqrt{x^2+1}}\right)^2=\frac{1}{x^2+1}$$

I derive this every time because I have a really bad memory but I think that you are supposed to memorise

$$\boxed{\frac{d}{dx}tan^{-1}x=\frac{1}{1+x^2}}$$

.Melody Oct 19, 2014

#19**+10 **

This one can be done without L'Hopital's Rule

$$\\lim \limits_{x\rightarrow 1}\;\;\frac{\sqrt{5-x}-2}{\sqrt{10-x}-3}\\\\

=lim \limits_{x\rightarrow 1}\;\;\frac{(\sqrt{5-x}-2)(\sqrt{10-x}+3)(\sqrt{5-x}+2)}{(\sqrt{10-x}-3)(\sqrt{10-x}+3)(\sqrt{5-x}+2)}\\\\

=lim \limits_{x\rightarrow 1}\;\;\frac{(5-x-4)(\sqrt{10-x}+3)}{(10-x-9)(\sqrt{5-x}+2)}\\\\

=lim \limits_{x\rightarrow 1}\;\;\frac{(1-x)(\sqrt{10-x}+3)}{(1-x)(\sqrt{5-x}+2)}\\\\

=lim \limits_{x\rightarrow 1}\;\;\frac{(\sqrt{10-x}+3)}{(\sqrt{5-x}+2)}\\\\

=\frac{(\sqrt{10-1}+3)}{(\sqrt{5-1}+2)}\\\\

=\frac{3+3}{2+2}\\\\

=\frac{6}{4}\\\\

=\frac{3}{2}\\\\$$

Melody Oct 20, 2014

#20**+5 **

Very nice "trick," Melody!!!....mutiplying by two conjugates !!!....I'll have to add that one to my "tool chest"

L'Hopital's is faster......but this is, somehow, more "mathematical"

CPhill Oct 20, 2014

#21**0 **

Thanks Chris,

Do you want me to prove L'Hopital's Rule is valid for a question like this ?

Melody Oct 20, 2014

#22**0 **

Nah...that's OK......I couldn't do that much thinking this early in the morning......!!!

CPhill Oct 20, 2014

#23**+10 **

Prove L'Hopital's rule for the following special case.

$$\\$Prove $\displaystyle\lim_{x\rightarrow a}\frac{f(x)}{g(x)}\; = \;\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\quad\\\\\\ $When $ g(a)=f(a)=0 $ and $ f'(a) $ and $ g'(a)$ exist $ \\\\\\ \displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\;\frac{\displaystyle\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}} {\displaystyle\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}\\\\\\\

\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\displaystyle\lim_{x \rightarrow a}\;\left(\frac{\frac{f(x)-f(a)}{x-a}} {\frac{g(x)-g(a)}{x-a}}\right)\\\\\\ \\\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\displaystyle\lim_{x \rightarrow a}\;\left(\frac{f(x)-f(a)}{ g(x)-g(a)}\right)\\\\\\

\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\displaystyle\lim_{x \rightarrow a}\;\left(\frac{f(x)-0}{ g(x)-0}\right)\\\\\\

\displaystyle\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\;=\displaystyle\lim_{x \rightarrow a}\;\frac{f(x)}{ g(x)}$$

Melody Oct 20, 2014