Lim sq.rt x[sqrt x - sqrt (x-a)] x->infinity

[Laven]

lim sq.rt x[sqrt x - sqrt (x-a)] x-->infinity

I'm whacked outta solving these problems just hanging around for more than 2 hours for each question but can't solve it.I think it's out of my capability.LOL.
I hope you will solve these problems.I need your help.The questions are...
1) lim sq.rt x[sqrt x - sqrt (x-a)]
x-->infinity

2) lim [tan x - sin x]/x^3
x--> 0

Thanks in advance.

 

[dx]

You must show your attempt to receive help.

 

[HallsofIvy]

You say you've been working on it for two hours (which is not really that much) so show what you have done in those two hours.

 

[Laven]

Yea you both are right so that i can catch where my mistake is but it's so tough to write mathematical notations overhere.Could you please give me a hint how can i write the mathematical notations?

 

[Office_Shredder]

https://www.physicsforums.com/showthread.php?t=8997

It's a little bit of work to learn how to do latex, but then you can ask all your questions here much easier

 

[Laven]

thanks office_shredder i got the idea to use latex.

Actually,for 1st question, i tried is a bit different one but similar to tht ones.The above questions seems complicated rather than i did.I gave yo because i get more idea to solve the question i tried.

Ok the question is [similar to tht one]

\lim_{x\rightarrow \infty} \sqrt{3x}-\sqrt{x-5}

= \lim_{x\rightarrow \infty} (\sqrt{3x}-\sqrt{x-5}) X (\sqrt{3x}+\sqrt{x-5})

 

[HallsofIvy]

thanks office_shredder i got the idea to use latex.

Actually,for 1st question, i tried is a bit different one but similar to tht ones.The above questions seems complicated rather than i did.I gave yo because i get more idea to solve the question i tried.

Ok the question is [similar to tht one]

\lim_{x\rightarrow \infty} \sqrt{3x}-\sqrt{x-5}

= \lim_{x\rightarrow \infty} (\sqrt{3x}-\sqrt{x-5}) \times (\sqrt{3x}+\sqrt{x-5})

No, you cannot just multiply by something without changing the value. You can multiply and divide by the same thing:
\left(\sqrt{3x}- \sqrt{x-5}\right)\frac{\sqrt{3x}+ \sqrt{x- 5}}{\sqrt{3x}+ \sqrt{x- 5}}
= \frac{3x- (x- 5)}{\sqrt{3x}+ \sqrt{x- 5}}= \frac{2x+ 5}{\sqrt{3x}+ \sqrt{x- 5}}

Now, to take the limit as x goes to infinity, divide both numerator and denominator by x, remembering that x will become x² inside the square roots.

 

[Laven]

Yea i know tht hallsofivy actually tht was my true mistake because i was just trying using latex & i can't get it well.

actually.the process I'm going to do is tht for the question
<br /> (\sqrt{3x}-\sqrt{x-5})<br />

is...
I multiplied it by conjugate on both denominator and numerator sides & i got
\frac{2x+5}{\sqrt{3x}+\sqrt{x-5}}
then i did 1st derivative on both numerator and denominator i got,

\frac{4\sqrt{x(x-5)}}{\sqrt{3(x-5)}+\sqrt{x}}

Again on 2nd derivative I got,
\frac{4x-10}{\sqrt{3x}+\sqrt{x-5}}

So,i can't converge this answer.Anyone have idea to solve this question.Where am i wrong?could yo please point it out?

 

[Laven]

For the 2nd question i.e \lim_{x\rightarrow0}(tan{x}-sin{x})
1stly i find its derivative [on both sides numerator and denominator] i got,
\frac{1-cos^3{x}}{3x^2cos^2{x}}
After further simplification with this i can't reach to the final answer.

Again on next way i change the trigonometric terms into half angle rule using formulae either way i can't get it.
Is there next way to solve this problem?

 

[slider142]

For the 2nd question i.e \lim_{x\rightarrow0}(tan{x}-sin{x})
1stly i find its derivative [on both sides numerator and denominator] i got,
\frac{1-cos^3{x}}{3x^2cos^2{x}}
After further simplification with this i can't reach to the final answer.

Again on next way i change the trigonometric terms into half angle rule using formulae either way i can't get it.
Is there next way to solve this problem?

Since the denominator is a 3rd degree monomial, it is 0 as x approaches 0 up to the 3rd derivative, which should be a constant. Thus, you should look for the limit at the third application of L'Hopital's rule. Take care that the numerator remains 0 at each step as well, otherwise it would be invalid to apply L'Hopital's rule.