Find limx-> 0 [(ex-1-x)/(x*sin(x))]

[JohanM]

Homework Statement

Find lim_{x-> 0} [(e^x-1-x)/(x*sin(x))]

The Attempt at a Solution

I tried using the Maclaurin series of e^x and sin(x) to get

[1+x+x²/2!+x³/3!+ ... -1 -x]/[x²-x⁴/3!+x⁶/5!-...]

but I don't know how to achieve the final answer of 1/2

 

[Sethric]

Have you considered trying L'Hospital's Rule?

 

[JohanM]

Have you considered trying L'Hospital's Rule?

I did try using L'Hospital's rule, only to get
lim_x->0 (e^x-1)/(sin(x)+x*cos(x))

Trying to use L'Hospital's repeatedly only makes the denominator more and more complicated.
Are you suggesting using L'Hospital's rule after replacing e^x and sin(x) with their Maclaurin series?

EDIT: Nevermind, I see what you mean. Using L'Hospital's rule twice yields
lim_x->0 e^x/(2*cos(x)-x*sin(x)) = 1/2

 

[vela]

Homework Statement

Find lim_{x-> 0} [(e^x-1-x)/(x*sin(x))]

The Attempt at a Solution

I tried using the Maclaurin series of e^x and sin(x) to get

[1+x+x²/2!+x³/3!+ ... -1 -x]/[x²-x⁴/3!+x⁶/5!-...]

but I don't know how to achieve the final answer of 1/2

If you simplify the numerator, you get

\lim_{x\to 0} \frac{\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots}{x^2-\frac{x^4}{3!}+\cdots}

Now get rid of the common factor of x² from the top and bottom.