Difficult Trig Limit

**NDHS** · November 23rd 2009, 08:02 PM

I seem to be having a tough time with this one limit problem:

Determine the limit of: $\lim_{x \rightarrow 0} \frac{tanx-sinx}{x^3}$

I've tried doing multiple simplifications but I have the problem of constantly getting a $\lim_{x \rightarrow 0} \frac{1-cosx}{x}=0$ which can't be right, as wolfram-alpha and the answer book is telling me that the answer is $\frac{1}{2}$ , which won't come from the zero multiplication!

Don't suppose I could snag a push in the right direction? thanks!

**Calculus26** · November 23rd 2009, 08:27 PM

If you can know L'Hopital's rule

lim [tan(x)-sin(x)]/x^3 = lim sin(x)/x lim (sec(x)-1)/x^2

= 1 lim[sec(x)tan(x)/2x

= lim [sec^2(x)tan(x) +sec^3(x)]/2

= 1/2

**Chris L T521** · November 23rd 2009, 08:27 PM

Originally Posted by NDHS

I seem to be having a tough time with this one limit problem:
I've tried doing multiple simplifications but I have the problem of constantly getting a $\lim_{x \rightarrow 0} \frac{1-cosx}{x}=0$ which can't be right, as wolfram-alpha and the answer book is telling me that the answer is $\frac{1}{2}$ , which won't come from the zero multiplication!

Don't suppose I could snag a push in the right direction? thanks!

Note that $\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}=\lim_{x\to 0}\frac{\sec x-1}{x^2}$

Now by L'Hôpital's rule, we have $\lim_{x\to 0}\frac{\sec x\tan x}{2x}\rightarrow\frac{0}{0}$ .

Applying the rule one more time yields $\lim_{x\to 0}\frac{\sec^3x+\sec^2x\tan x}{2}=\frac{1}{2}$ .

Does this make sense?

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