Mr. Snookums said:
I know how to solve a regular question such as x^x, but how would I find:
lim[1-(4/x)]^x
x->inf
My study guide says that I have to use Ln and then L'Hopital's Rule, and I can see how that would work, but what happens to the limit?
Yes, you can use ln, and then L'Hopital's rule. However, there's a simplier way to do this.
Since:
\lim_{x \rightarrow + \infty} \left( 1 + \frac{1}{x} \right) ^ x = \lim_{x \rightarrow - \infty} \left( 1 + \frac{1}{x} \right) ^ x = e
So ingeneral:
\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e
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To solve the problem, you must try to make the expression have the form above. So, we have:
\lim_{x \rightarrow \infty} \left( 1 - \frac{4}{x} \right) ^ x = \lim_{x \rightarrow \infty} \left( 1 - \frac{1}{\frac{x}{4}} \right) ^ x = \lim_{x \rightarrow \infty} \left( 1 + \frac{1}{\frac{- x}{4}} \right) ^ x
= \lim_{x \rightarrow \infty} \left[ \left( 1 + \frac{1}{\frac{- x}{4}} \right) ^ \frac{-x}{4} \right] ^ {-4}
Now, can you go from here? :)
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If you want to use
L'Hopital's rule, you can let:
y = \lim_{x \rightarrow \infty} \left( 1 - \frac{4}{x} \right) ^ x
\Rightarrow \ln y = \ln \lim_{x \rightarrow \infty} \left( 1 - \frac{4}{x} \right) ^ x = \lim_{x \rightarrow \infty} \ln \left( 1 - \frac{4}{x} \right) ^ x = \lim_{x \rightarrow \infty} \left[ x \ln \left( 1 - \frac{4}{x} \right) \right] = \lim_{x \rightarrow \infty} \left[ \frac{\ln \left( 1 - \frac{4}{x} \right) }{\frac{1}{x}} \right] = z
Now, you can apply
L'Hopital's rule here. The RHS is the
Indeterminate form 0 / 0.
After applying
L'Hopital's Rule, we can obtain y by the fact that y = e
z.
Can you go from here? :)
