Hope I can do this

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

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.  

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.

yes its very clear .. 

what about the rest of the questions ? 

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.

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

https://www.desmos.com/calculator/k532qggqyb

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 

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.     

2nd one

$$\\\displaystyle\lim_{x\rightarrow1}\;\frac{x-1}{lnx}\\\\
$on first inspection this is 0/0 so you can use L'Hopital's Rule$\\\\\\
=\displaystyle\lim_{x\rightarrow1}\;\frac{1}{1/x}\\\\
=\frac{1}{1/1}\\\\
=1$$

so there are no way to solve these problems without L'Hopital's Rule  right  ? 

Beats me - Ask Alan   LOL  

Thank you alan but what do you mean by beats me ? 

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.    

Also how do you get this ?

which I highlighted

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

  (x² - 1)/(x² - 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

.

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 - z²/2 - z³/3 - z⁴/4 - ...

so -z/ln(1 - z) = 1/(1 + z/2 + z²/3 + z³/4 + ...)

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!

.

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

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

thats nice .. thank you alan and melody .. 

is my answer here correct ? 

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

$$\\lim \limits_{x\rightarrow 1}\;\;\frac{\sqrt{5-x}-2}{\sqrt{10-x}-3}\\\\
=lim \limits_{x\rightarrow 1}\;\;\frac{(\sqrt{5-x}-2)(\sqrt{10-x}+3)(\sqrt{5-x}+2)}{(\sqrt{10-x}-3)(\sqrt{10-x}+3)(\sqrt{5-x}+2)}\\\\
=lim \limits_{x\rightarrow 1}\;\;\frac{(5-x-4)(\sqrt{10-x}+3)}{(10-x-9)(\sqrt{5-x}+2)}\\\\
=lim \limits_{x\rightarrow 1}\;\;\frac{(1-x)(\sqrt{10-x}+3)}{(1-x)(\sqrt{5-x}+2)}\\\\
=lim \limits_{x\rightarrow 1}\;\;\frac{(\sqrt{10-x}+3)}{(\sqrt{5-x}+2)}\\\\
=\frac{(\sqrt{10-1}+3)}{(\sqrt{5-1}+2)}\\\\
=\frac{3+3}{2+2}\\\\
=\frac{6}{4}\\\\
=\frac{3}{2}\\\\$$

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"

Thanks Chris,

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

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

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

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

I was already doing it before you answered LOL.  :)

OK...that's not as bad as I thought it would be......a couple of G & T's later on would probably help "reinforce" the concept!!!!