Question

Use L Hopital's Rule to evaluate the following limits:

(I'm having problems with three questions based on the instruction and I'll list them one by one. Really need help pleade ^.^;)

Answer 1

what do u mean??

Answer 2

Wait, I'll give the first question out ^.^;

Answer 3

\[\lim_{x \rightarrow 0} 2x/(x+7\sqrt{x})\]

Answer 4

btw I meant please, not pleade ^.^

Answer 5

whats the correct answer @lightchaste ?

Answer 6

@fatinatikah The answer is 0. I got it right by using the conjugate method but I want to know how the steps are supposed to be made by using L Hopital's Rule

Answer 7

u sure the answer is 0?

Answer 8

can i know your steps?

Answer 9

yes the answer is 0

Answer 10

can i know the working?

Answer 11

One hint. Take LCM in the denominator :)

Answer 12

differentiating both denominator and nominator gives
2 in numerator
and {1+3.5/(x^1/2)}

Answer 13

yep the answer is 0
Using the conjugate method:
\[\lim_{x \rightarrow 0}2x/(x+7\sqrt{x})\]\[\lim_{x \rightarrow 0} 2x(x-7\sqrt{x})/[(x-1\sqrt{x})(x+7\sqrt{x})]\]\[\lim_{x \rightarrow 0}2x(x-7\sqrt{x})/(x ^{2}-49x)\]
\[\lim_{x \rightarrow 0} 2(x-7\sqrt{x})/(x-49)\]
Replace it with the limit you'll get 0

Answer 14

put 0 in equation, in denominator, it becomes
\[(1+3.5/0)\]

Answer 15

3.5/0 makes it infinity ,
then infinity +1 =infinity
hence the final equation is
2/infinity
and anything divided by infinity gives 0

Answer 16

did u get it man?

Answer 17

Doesn't 3.5/0 become indeterminate?

Answer 18

indeterminate forms applies to whole equation not a part of it, like in this case

Answer 19

BTW @fatinatikah, I fumbled on the second line of my equation, it's not \[1\sqrt{x}\] it's \[7\sqrt{x}\] ^.^

Answer 20

@wasiqss is right :

Also it doesn't matter

Because 7 * 0 =1 *0=0

Answer 21

Because 7 * 0 =1 *0=0
-->This is in reference to mistake in question 7 sqrt(x) instead o f 1sqrt(x)

Answer 22

ahhhh now I understand :)

Answer 23

@shivam_bhalla owh lol right, but since I'm using conjugate it wouldn't make sense in the working even though the answer is right :)

Answer 24

Question 2 coming up in 20 mins ay :D

Answer 25

i will be gone till then :P

Answer 26

haha lol sorry, gotta do something urgent at the moment :(

Answer 27

aww its fine

Answer 28

But you know what, I'll just type it in quickly
\[\lim_{r \rightarrow 1}a(r ^{n}-1)/(r-1)\]

the answer is an, but I got an-1 :(

Answer 29

Simple. Tell me what is differential of r^n with respect to n.

Answer 30

Well just for the sake of having a reponse/reply, the third question is:
\[\lim_{x \rightarrow \infty} (x -\sqrt{x ^{2}+x})\]

Answer 31

urm..nr^(n-1)?

Answer 32

yes. Applying l'hopital rule,

(d/dn) ( denominator) = 1

(d/dn) ( a *r^n) = a * (d/dn)( r^n) = a*n*r^(n-1)

substitute r=1. So 1^(n-1)=1

a*n is the answer

Answer 33

owh so is my way incorrect?
\[\lim_{r \rightarrow 1} (ar ^{n}-1)/(r-1)\]\[\lim_{r \rightarrow 1}(anr ^{n}-1)/1 \] (after being differentiated)
\[an -1\]

Answer 34

Differentiation of constant i.e -1 in your case is zero :)
and Sorry for replying late because I was busy at another problem :)

Answer 35

lol I typed the first line wrong. it's supposed to be (ar^n -r)

Answer 36

Well since this is going to be a pretty long post ill just close it >.>