Question

Limits question.. Got stuck in between :(

The question is :
\(\lim_{x \rightarrow 1} \cfrac{x^2 - \sqrt{x}}{\sqrt{x} - 1}\)

Answer 1

multiply by conjugate

Answer 2

I tried to apply LH rule and rationalizing too.

Answer 3

\(\lim_ {x\rightarrow 1} \cfrac{(x^2-\sqrt{x})(\sqrt{x}+1)}{x - 1}\)

(After rationalizing)

Answer 4

Foil then divide each term by x^2?

Answer 5

What if I want to apply LH rule here?

Answer 6

It's grody, but assuming you have 0/0 sure

Answer 7

I'd still foil first

Answer 8

then LH isn't bad

Answer 9

I am getting 3/4 , is it right?

Answer 10

I would use a substitution u=sqrt(x) then maybe take. Care of the factoring.

Answer 11

wolf got just 3

Answer 12

I like freckles idea

Answer 13

By LH rule
\[
\frac{\text{Num}'(x)}{\text{Den}'(x)}=2 \sqrt{x} \left(2 x-\frac{1}{2
\sqrt{x}}\right)=(4 x^{3/2}-1)\to 3 \text { when } x\to 1
\]

Answer 14

Okay, so here is what I did :

\(\lim _ {x\rightarrow 1 } \cfrac{(x^2 - \sqrt{x})(\sqrt{x}+1)}{x-1} \\
\lim _ {x\rightarrow 1 }[ ({2x - \cfrac{1}{2} \times \cfrac{1}{\sqrt{x} } )\times (\sqrt{x} + 1) ]+[ ({x^2 - \sqrt{x}})+{ \cfrac{1}{2\sqrt{x}}] } }\)

Answer 15

Oops, have I just made a mistake in that? I don't think, I have done anything right above.. :(

Answer 16

The limit is 3

Answer 17

\(\lim _ {x\rightarrow 1 } \cfrac{(x^2 - \sqrt{x})(\sqrt{x}+1)}{x-1} \\
\lim _ {x\rightarrow 1 }[ ({2x - \cfrac{1}{2} \times \cfrac{1}{\sqrt{x} } )\times (\sqrt{x} + 1) ]+[ ({x^2 - \sqrt{x}})\times { \cfrac{1}{2\sqrt{x}}] } }\)

and then I put x = 1 , so, I get 3 as well :)

Answer 18

I think, I made it much complicated... At first, I didn't apply this concept while applying LH Rule :

\(\cfrac{d}{dx} (uv) = u'v + uv'\)
and so, I was getting 3/4 ...

Answer 19

No need to rationalize first

Answer 20

Okay.. , so directly apply LH rule?

Answer 21

I got it... If I apply LH rule at the first step, it would have been easier. Thank you everyone for making it clear for me.. :)

Answer 22

YW

Answer 23

\[\lim_{x \rightarrow 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}\]
let \[u=\sqrt{x}\]
\[\text{so x goes to 1, then } u \text{ goes to } \sqrt{1}=1\]
So we have
\[\lim_{u \rightarrow 1}\frac{u^4-u}{u-1}\]
So \[u^4-u=u(u^3-1)=u(u-1)(u^2+u+1)\]
Now you are able to get rid of the the discontinuity

Answer 24

Okay, so, (u-1) will cancel out and I will just put u = 1 in the left expression and thus, I'll get 3. :)

Nice work @freckles , it is really interesting to get to know about more methods to solve a single question. :)