Question

Evaluate the limit:

Answer 1

\[\Large \lim_{x \rightarrow 0^+} ~sinx~lnx\]

Answer 2

\[ \lim_{x \rightarrow 0^+} ~\sin x~\ln x = \lim_{x \rightarrow 0^+} \dfrac{\ln x}{\csc x} = \lim_{x \rightarrow 0^+} \dfrac{1/x}{-\csc x \cot x} = \lim_{x \rightarrow 0^+} \dfrac{\sin x }{x} (- \tan x) \]

Answer 3

take the limit

Answer 4

I knew I had to rewrite it somehow, just wasn't sure if rewriting sine was it. Thanks :)

Answer 5

@ganeshie8 - Can you please explain, how did you come up to the third step? Did you apply L'Hopital's Rule?

Answer 6

\[hint: \lim_{x \rightarrow 0^{+}}\frac{ sinx }{ x }=1\]

Answer 7

sinx = 1/cscx

Answer 8

np :)

below all are convertible to 0/0 indeterminate form for L'Hospital rule :

\(0 \times \infty, ~ 0^0, ~ 1^{\infty}, ~ \infty^0\)

Answer 9

yes @mathslover .... @iambatman has it :)

Answer 10

Oh okay! Fine.. Thanks @ganeshie8