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

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The discussion revolves around solving two limit problems: the first limit involves the expression lim sq.rt x[sqrt x - sqrt (x-a)] as x approaches infinity, while the second limit concerns lim [tan x - sin x]/x^3 as x approaches 0. Participants express frustration with the complexity of the problems and share their attempts at solutions, including using derivatives and the conjugate method. A suggestion is made to use LaTeX for clearer mathematical notation, which some users find helpful. The conversation emphasizes the importance of correctly applying L'Hôpital's rule for the second limit to achieve the correct solution.
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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.
 
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You must show your attempt to receive help.
 


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.
 


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?
 


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})
 


Laven said:
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.
 


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?
 


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?
 
  • #10


Laven said:
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.
 
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