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# Hope I can do this

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Hope I can do this Oct 18, 2014

#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   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. Oct 19, 2014
#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
#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.

Oct 19, 2014
#4
0

yes its very clear ..

what about the rest of the questions ?

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.

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

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

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 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. Oct 19, 2014
#9
+5

2nd one

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

.
Oct 19, 2014
#10
0

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

Oct 19, 2014
#11
+5

Beats me - Ask Alan   LOL Oct 19, 2014
#12
0

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

Oct 19, 2014
#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. Oct 19, 2014
#14
0

Also how do you get this ?

which I highlighted Oct 19, 2014
#15
+10

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

.

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 - 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!

.

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

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

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Oct 19, 2014
#18
0

thats nice .. thank you alan and melody ..

is my answer here correct ? Oct 19, 2014
#19
+10

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

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

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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"   Oct 20, 2014
#21
0

Thanks Chris,

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

Oct 20, 2014
#22
0

Nah...that's OK......I couldn't do that much thinking this early in the morning......!!!   Oct 20, 2014
#23
+10

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

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

Oct 20, 2014
#24
+5   