Compute the limit x→0 (t +1/t) [ (4 -t)^(3/2) - 8]

[hndalama]

limit x→0 (t +1/t) [ (4 -t)^(3/2) - 8]

I try to solve this by using the limit property that if lim f(x) = L and lim g(x) = M then lim f(x)g(x) = LM
to solve the limit for (t + 1/t) I rewrite it as (t²+1)/t then using Lhopital's rule it becomes 2t.
then I evaluate the limits at 0, leaving me with L =0 and M = 0 so my answer is 0.
The answer given is -3. How do they get that answer? and what is wrong with my method?

 

[ksdhart2]

Sometimes splitting the limit apart into two limits can be helpful, but this isn't one of those cases. Where you went wrong is that you can't actually use L'Hopital's Rule on f(x). Let's take the original function and plug in zero and see what happens:

\(\displaystyle \displaystyle \lim _{t\to 0}\left(\frac{t^2+1}{t}\right) = \frac{1}{0}\)

Oops! 1/0 is not one of the forms that we can use L'Hopital's Rule on. It only works on \(\displaystyle \dfrac{0}{0}\) or \(\displaystyle \dfrac{\pm \infty}{\pm \infty}\)

Instead, let's combine the two fractions:

\(\displaystyle \displaystyle \lim _{t\to 0}\left(\left(\frac{t^2+1}{t}\right)\left(\left(4-t\right)^{\frac{3}{2}}-8\right)\right)=\lim _{t\to 0}\left(\frac{t^2+1\left(\left(4-t\right)^{\frac{3}{2}}-8\right)}{t}\right)\)

From there, your best bet would probably be to multiply by the conjugate.

 

[hndalama]

Sometimes splitting the limit apart into two limits can be helpful, but this isn't one of those cases. Where you went wrong is that you can't actually use L'Hopital's Rule on f(x). Let's take the original function and plug in zero and see what happens:

\(\displaystyle \displaystyle \lim _{t\to 0}\left(\frac{t^2+1}{t}\right) = \frac{1}{0}\)

Oops! 1/0 is not one of the forms that we can use L'Hopital's Rule on. It only works on \(\displaystyle \dfrac{0}{0}\) or \(\displaystyle \dfrac{\pm \infty}{\pm \infty}\)

Instead, let's combine the two fractions:

\(\displaystyle \displaystyle \lim _{t\to 0}\left(\left(\frac{t^2+1}{t}\right)\left(\left(4-t\right)^{\frac{3}{2}}-8\right)\right)=\lim _{t\to 0}\left(\frac{t^2+1\left(\left(4-t\right)^{\frac{3}{2}}-8\right)}{t}\right)\)

From there, your best bet would probably be to multiply by the conjugate.

ah, I didn't realise that LHopital's rule can only be used on the indeterminate forms. thanks for your help