Question

Questions from Unit 1: Calculus

Answer 1

I feel like this is easy but I don't remember how to do it @smokeybrown

Answer 2

\[\frac{ \frac{ 1 }{ x }-1 }{ x -1 } = ? \]

like a first step how you simplifie this fraction ?

Answer 3

\[\lim_{x \rightarrow 1}\frac{ \frac{ 1 }{x }-1 }{ x-1 }\]
\[=\lim_{x \rightarrow 1}\frac{ 1-x }{ x(x-1) }\]
\[=\lim_{x \rightarrow1}\frac{ -(x-1) }{ x(x-1) }\]
\[=\lim_{x \rightarrow 1}(\frac{ -1 }{ x })\]

Answer 4

Jhonyy's hint is how you can get to the correct answer.

One thing that I would want to add is the limit properties and to not get confused

\[\lim_{x \rightarrow a}\left(\dfrac{f(x)}{g(x)}\right)= \large\dfrac{\lim_{x \rightarrow a}f(x)}{\lim_{x \rightarrow a}g(x)}\]

But only if
\[\lim_{x \rightarrow a}g(x) \neq 0\]

That's why you have to simplify the fraction. Because the limit of x-1 as x approaches 0 would be 0 so you cannot apply this property

Answer 5

Thanks AZ, Jhonyy9 and surij!!!
So another way I thought maybe I could do it is that
\[\lim_{x \rightarrow 1}\frac{ \frac{ 1 }{x }-1 }{ x-1 }\]
so if i plug in 1 for x I get 0/0 which is indeterminate form, so then \[\lim_{x \rightarrow 1}\frac{ \frac{ 1 }{x }-1 }{ x-1 } =\lim_{x \rightarrow 1}-1/x^2\] and when i plug in 1 I get -1. And then I could maybe match all the other answer choices and see which evaluates to -1 which is A

Answer 6

\[\lim_{x \rightarrow 1}\frac{ \frac{ 1 }{x }-1 }{ x-1 } =\lim_{x \rightarrow 1}-1/x^2\] i just took the derivative of top and bottom, derivative of x-1 is just 1

Answer 7

L'Hospitals Rule

Answer 8

L'Hopital's rule works too. Well done!

Answer 9

thanks!