Question

Anyone up for a little challenge with limits?

Answer 1

\[\lim_{x \rightarrow -1} \frac {x + 1}{\sqrt[4]{x+17} - 2}\]

Answer 2

\[\lim_{x \rightarrow -1}(x+1)(\sqrt[4]{x+17}+2)/\sqrt{x+17}-4\]

Answer 3

\[\lim_{x \rightarrow -1}(x+1)(\sqrt[4]{x+17}+2)(\sqrt{x+17}+4)/x+17-16\]

Answer 4

so the answer is 32

Answer 5

Nice job! Double rationalization works, but I was looking for a cleaner method :P

Answer 6

Want me to post the cleaner method or do you wanna figure it out? =P

Answer 7

sure

Answer 8

Alright, you have to transform the limit with a substitution.

Answer 9

\[\lim_{x \rightarrow -1} \frac {x + 1}{\sqrt[4]{x+17} - 2}\]\[u = \sqrt[4]{x+17}, x = u^4 - 17\]As x --> -1, u ---> 2\[\lim_{u \rightarrow 2} \frac {u^4 - 16}{u - 2} = \lim_{u \rightarrow 2} (u + 2)(u^2 + 4) = 32\]

Answer 10

nice one i was thinking you were using Taylor series

Answer 11

Its an elegant method my calc teacher taught us in the beginning of the year, I love it =)

I didn't know that you could use taylor series to do limits. Well, we started taylor series last Friday, so...

Can you show me how you would do it w/ taylor?

Answer 12

\[\sqrt[4]{x+17}-2 =\sum_{n=0}^{\infty}(\prod_{k=1}^{n}(0.25-k))(-1+17)^{0.25-n}(x+1)^{n}/n!- 2\]
\[=\sum_{n=1}^{\infty}(\prod_{k=1}^{n}(0.25-k))(-1+17)^{0.25-n}(x+1)^{n}/n!+2- 2\]
So that we have:
\[\lim_{x \rightarrow -1}(x+1)/(\sqrt[4]{x+17}-2)\]
\[=\lim_{x \rightarrow -1}(x+1)/(\sum_{n=1}^{\infty}(\prod_{k=1}^{n}(0.25-k))(-1+17)^{0.25-n}(x+1)^{n}/n!)\]
\[=\lim_{x \rightarrow -1}1/(\sum_{n=1}^{\infty}(\prod_{k=1}^{n}(0.25-k))(-1+17)^{0.25-n}(x+1)^{n-1}/n!)\]
\[=\lim_{x \rightarrow -1}1/(\sum_{n=2}^{\infty}(\prod_{k=1}^{n}(0.25-k))(-1+17)^{0.25-n}(x+1)^{n-1}/n!+0.25*16^{-0.75})\]
=32
But this method is way far from elegant lol.

Answer 13

lol, wow, an infinite sum & product. Not pretty at all.