Question

lim x->0^+ (sinx)/(x+sqrootx)

Answer: 0

My question, why 0? Shouldn't it be infinity?

Answer 1

if x = 0
this equals 0 / 0
so use l'hopitals rule
differentiate top and bottom

Answer 2

i'll take a look at it - but limits are not my strong point

Answer 3

So I used l'hopitals rule and I keep getting 0/0

Answer 4

heres the man - zarkon

Answer 5

:D

Answer 6

look at it like this...

\[\frac{\sin(x)}{x+\sqrt{x}}=\frac{1}{\frac{x+\sqrt{x}}{\sin{x}}}=\frac{1}{\frac{x}{\sin(x)}+\frac{\sqrt{x}}{\sin{x}}}\]

Answer 7

\[\lim_{x \rightarrow 0^+}\frac{\sin(x)}{x+\sqrt{x}} \cdot \frac{x- \sqrt{x}}{x-\sqrt{x}}=\lim_{x \rightarrow 0^+}\frac{\sin(x)(x-\sqrt{x})}{x^2-x}\]
\[\lim_{x \rightarrow 0^+}\frac{\sin(x)(x-\sqrt{x})}{x(x-1)}=\lim_{x \rightarrow 0^+}\frac{\sin(x)}{x} \cdot \lim_{x \rightarrow 0^+} \frac{x-\sqrt{x}}{x-1}\]

Answer 8

choose my way :)

Answer 9

lol

Answer 10

haha thanks :D, so at the end what should i replace x as? A number close to 0? The answer they give me is 0 :S

Answer 11

\[\lim_{x \rightarrow 0^+}\frac{\sin(x)}{x}=1\]
\[\lim_{x \rightarrow 0^+}\frac{x-\sqrt{x}}{x-1}=\frac{0-\sqrt{0}}{0-1}=0\]

the answer is 0*1
which can be simplify to .....
you fill in this blank lol

Answer 12

Ohhhhhhhh, I get it now.. I guess I should be more careful next time. Thanks a lot , I appreciate it.