Question

How to do this L'Hospital Problem?

http://i.imgur.com/0gk7URt.png

Answer 1

using white ink on a white paper
since it's blank :/

Answer 2

@jdoe0001 just fixed it fam

Answer 3

pls halp

Answer 4

looking

Answer 5

<3

Answer 6

hmmm what's hmmm the answer if any provided?

Answer 7

The answer listed is correct
(0)

Answer 8

hmm

Answer 9

one sec

Answer 10

\(\bf \lim\limits_{x\to0^{+}}\ sin(x)ln(3x)
\\ \quad \\
{\color{brown}{ sin(x)\iff\frac{1}{[sin(x)]^{-1}} }}\ thus\ \cfrac{1}{[sin(x)]^{-1}}\cdot ln(3x)
\\ \quad \\
\cfrac{ln(3x)}{[sin(x)]^{-1}}\implies \underline{LH}\implies \cfrac{\frac{d}{dx}[ln(3x)]}{\frac{d}{dx}\left[ \frac{1}{sin(x)}\right]}
\\ \quad \\
\cfrac{\frac{\cancel{3}}{\cancel{3}x}}{-\frac{cos(x)}{sin^2(x)}}\impliedby \textit{bottom one obtained by using quotient rule}
\\ \quad \\
\cfrac{\frac{1}{x}}{-\frac{cos(x)}{sin^2(x)}}\implies \cfrac{-sin^2(x)}{{\color{brown}{ x}} cos(x)}\implies -\cfrac{sin(x)}{{\color{brown}{ x}}}\cdot \cfrac{sin(x)}{cos(x)}
\\ \quad \\
-1\cdot \cfrac{sin(0)}{cos(0)}\implies -1\cdot \cfrac{0}{1}\implies -1\cdot 0\)

Answer 11

anyhow.... the last part, after using the LH rule is
limit of A * B which is the same as (Lim A) * (Lim B)
which ends up as -1 * 0

Answer 12

You are bae, thank you

Answer 13

yw