Solving Limits: "x → k" | Math Test

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The discussion revolves around solving the limit expression \(\lim_{x \rightarrow k} \frac{x\sqrt{x} - k\sqrt{k}}{x^{4}-k^{4}}\). Participants suggest various methods, including L'Hôpital's rule and factoring the denominator. A key point made is to divide both the numerator and denominator by \(x - k\) to simplify the limit, ultimately leading to a form suitable for L'Hôpital's rule. It is emphasized that L'Hôpital's rule can only be applied when the expression results in a 0/0 or ∞/∞ form. The conversation concludes with alternative approaches, such as changing variables to facilitate the limit evaluation.
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Hey everyone,
I had a math test yesterday. It was pretty hard... This is one of the limits I wasn't able to do.

\lim_{x \rightarrow k} \frac {x\sqrt{x} - k\sqrt{k}}{x^{4}-k^{4}}

I tried the Hopital rule, I tried multiplying the whole expression with the denominator. I didn't get to anything better.

Anyone knows how to do this kind of limits ? Thank you!
 
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Hint: Divide numerator and denominator by x-k.
Then use
\lim_{x \rightarrow k} \frac {x^n-k^n}{x-k}=nk^{n-1}
for both num and den.
 
mohlam12 said:
Hey everyone,
I had a math test yesterday. It was pretty hard... This is one of the limits I wasn't able to do.

\lim_{x \rightarrow k} \frac {x\sqrt{x} - k\sqrt{k}}{x^{4}-k^{4}}

I tried the Hopital rule, I tried multiplying the whole expression with the denominator. I didn't get to anything better.

Anyone knows how to do this kind of limits ? Thank you!

Factorise the denominator. (x^4 - k^4) = (x-k)(x+k)(x^2+k^2). The rightmost two factors can be evaluated immediately at the limit, they become (2k) and (2k^2) respectively, yes ?

Then the limit becomes

\frac{1}{(2k)(2k^2)}\lim_{x \rightarrow k} \frac {x^{1.5} - k^{1.5}}{x-k}

Now observe that the limit that's left is of the form 0/0, and can be reduced by LH rule. Just differentiate numerator and denominator wrt x. Put x = k into that, simplify the algebra and you're left with an expression in k.
 
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Why can't L'Hopital's rule work from the start?
 
BerryBoy said:
Why can't L'Hopital's rule work from the start?

It can ! Stupid me. Orig poster, disregard my post and just differentiate numerator and denominator to get a single expression in x and set x = k.
 
If you use L'Hopital rule in \frac {x^{1.5} - k^{1.5}}{x-k}, you will get \frac {6 \sqrt{k} -6 \sqrt{x}}{4 \sqrt{kx} *(x-k)'}
Then what can we do ?
 
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mohlam12 said:
If you use L'Hopital rule in \frac {x^{1.5} - k^{1.5}}{x-k}, you will get \frac {6 \sqrt{k} -6 \sqrt{x}}{4 \sqrt{kx} *(x-k)'}
Then what can we do ?

You're not differentiating correctly, the k is a constant and vanishes from both numerator and denominator.
 
:bugeye: !
If only I knew when passing the test! :frown:
Thank you guys
 
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Please note that L'Hopital rule can only be used when your expression is equal to 0/0 or inf/inf. so you need to check your expression each time before you use the rule.
 
  • #10
Multiply both the denominator and the numerator by x\sqrt{x}+k\sqrt{k} and then simplify the fraction by x-k.


Daniel.
 
  • #11
Or change the variable u=x^4 and let a=k^4, then u \to a as x \to k and the limit becomes:

\lim_{u \to a} \frac{u^{3/8}-a^{3/8}}{u-a}
which is the derivative of f(u)=u^{3/8} at u=a.
 
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