Question

Limit help?

Answer 1

\[\lim_{x \rightarrow 0} \frac{ \frac{ 1 }{ x+1 }-1 }{ x }\]

Answer 2

what did you do so far?

Answer 3

I didn't do anything so far. I am at a complete loss to solve it. I've done other problems with radicals in the numerator, but not something like this.

Answer 4

Try getting rid of the complex fraction (the fraction contained inside the fraction)

Answer 5

well consider the top fraction. simplify it then see what happens

Answer 6

\(\bf \lim_{x \to 0} \cfrac{ \frac{ 1 }{ x+1 }-1 }{ x }
\\ \quad \\
\cfrac{ \frac{ 1 }{ x+1 }-1 }{ x }\implies \cfrac{\frac{(1)(1)-(x+1)(1)}{x+1}}{x}\implies \cfrac{\frac{\cancel{ 1 }-x\cancel{ -1 }}{x+1}}{x}
\\ \quad \\
\cfrac{-\cancel{ x }}{x+1}\cdot \cfrac{1}{\cancel{ x }}\implies ?\)

Answer 7

1 thank you!!

Answer 8

Welcome^_^