limit of a trigonometric function

PROBLEM:

$\displaystyle \lim_{x\to\0}\frac{sin(x)}{tan(x)+x}$

ATTEMPT:

I'm stumped on this one for two reasons. (1) I don't see any trigonometric identities that would be convenient to help evaluate the limit, and (2) when I transformed the expression in terms of sine and cosine (to make apparent the properties of limits), it doesn't seem to work either.

Here's what I've done:

$\displaystyle \lim_{x\to\0}\frac{sin(x)}{tan(x)+x}=\lim_{x\to\0} \frac{sin(x)cos(x)}{sin(x)+xcos(x)}$

If I attempted to use properties of limits to evaluate the expression in parts, but it's not very useful unless I could do something about the denominator:

$\displaystyle \lim_{x\to\0}(\frac{sin(x)}{x})(\frac{xcos(x)}{sin (x)+xcos(x)})$

Any ideas?

Re: limit of a trigonometric function

Quote:

Originally Posted by

**Lambin** PROBLEM:

$\displaystyle \lim_{x\to\0}\frac{sin(x)}{tan(x)+x}$

ATTEMPT:

I'm stumped on this one for two reasons. (1) I don't see any trigonometric identities that would be convenient to help evaluate the limit, and (2) when I transformed the expression in terms of sine and cosine (to make apparent the properties of limits), it doesn't seem to work either.

Here's what I've done:

$\displaystyle \lim_{x\to\0}\frac{sin(x)}{tan(x)+x}=\lim_{x\to\0} \frac{sin(x)cos(x)}{sin(x)+xcos(x)}$

If I attempted to use properties of limits to evaluate the expression in parts, but it's not very useful unless I could do something about the denominator:

$\displaystyle \lim_{x\to\0}(\frac{sin(x)}{x})(\frac{xcos(x)}{sin (x)+xcos(x)})$

Any ideas?

Since the top and bottom both go to 0, L'Hospital's Rule can be used.

$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{(x)}}{\tan{(x)} + x} &= \lim_{x \to 0}\frac{\frac{d}{dx} \left[ \sin{(x)} \right]}{\frac{d}{dx}\left[ \tan{(x)} + x \right] } \\ &= \lim_{x \to 0} \frac{\cos{(x)}}{\sec^2{(x)} + 1} \\ &= \lim_{x \to 0}\frac{1}{1^2 + 1} \\ &= \frac{1}{2} \end{align*}$

Re: limit of a trigonometric function

Quote:

Originally Posted by

**Prove It** Since the top and bottom both go to 0, L'Hospital's Rule can be used.

$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{(x)}}{\tan{(x)} + x} &= \lim_{x \to 0}\frac{\frac{d}{dx} \left[ \sin{(x)} \right]}{\frac{d}{dx}\left[ \tan{(x)} + x \right] } \\ &= \lim_{x \to 0} \frac{\cos{(x)}}{\sec^2{(x)} + 1} \\ &= \lim_{x \to 0}\frac{1}{1^2 + 1} \\ &= \frac{1}{2} \end{align*}$

Thanks for replying. We haven't yet covered L'Hopital's rule. Is there any way around it?

Re: limit of a trigonometric function

L'Hospital's Rule is the quickest way, but another is to try to use the well known limit $\displaystyle \displaystyle \lim_{x \to 0}\frac{\sin{(x)}}{x} = 1$.

$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{(x)}}{\tan{(x)} + x} &= \lim_{x \to 0}\frac{\sin{(x)}}{\frac{\sin{(x)}}{\cos{(x)}} + x} \\ &= \lim_{x \to 0}\frac{\frac{1}{x}\left[ \sin{(x)} \right]}{\frac{1}{x} \left[ \frac{\sin{(x)}}{\cos{(x)}} + x \right] } \\ &= \lim_{x \to 0}\frac{\frac{\sin{(x)}}{x}}{\frac{\sin{(x)}}{x} \cdot \frac{1}{\cos{(x)}} + 1} \\ &= \frac{1}{1\cdot 1 + 1} \\ &= \frac{1}{2} \end{align*}$

Re: limit of a trigonometric function

Quote:

Originally Posted by

**Lambin** Thanks for replying. We haven't yet covered L'Hopital's rule. Is there any way around it?

I once did a ten year experiment teaching *honor calculus* in which we allowed no use of L'Hopital's rule nor u-substitution.

Even if I do say so, we had a great success with the numbers of students went on to the PhD or MD from those classes.

So I say: never use the L'Hopital's rule if it can be avoided.

$\displaystyle \frac{\sin(x)}{\tan(x)+x}=\frac{\dfrac{\sin(x)}{x} }{\dfrac{\sin(x)}{x}\cdot\dfrac{1}{\cos(x)}+1}$

Re: limit of a trigonometric function

Quote:

Originally Posted by

**Plato** I once did a ten year experiment teaching *honor calculus* in which we allowed no use of L'Hopital's rule nor u-substitution.

Even if I do say so, we had a great success with the numbers of students went on to the PhD or MD from those classes.

So I say: never use the L'Hopital's rule if it can be avoided.

$\displaystyle \frac{\sin(x)}{\tan(x)+x}=\frac{\dfrac{\sin(x)}{x} }{\dfrac{\sin(x)}{x}\cdot\dfrac{1}{\cos(x)}+1}$

I refrain from using L'Hopital's rule whenever possible. It's a very useful tool, but it prevents me from seeing the equation for what it actually is. I'll keep this in mind. Thanks to both of you.

Re: limit of a trigonometric function

You must enjoy making life difficult on yourselves...

Re: limit of a trigonometric function

Quote:

Originally Posted by

**Prove It** You must enjoy making life difficult on yourselves...

On the contrary.

We want to know how it works not why it works.

That is the historical difference in *mathematics* and *science*.

L'Hopital's rule is the *why* but not really the *how*.

Re: limit of a trigonometric function

Quote:

Originally Posted by

**Plato** On the contrary.

We want to know how it works not why it works.

That is the historical difference in *mathematics* and *science*.

L'Hopital's rule is the *why* but not really the *how*.

I would say that its a matter of opinion. Both are equally legit strategies for finding the limit, both make sense, both are backed by proof and by intuition.

Re: limit of a trigonometric function

Quote:

Originally Posted by

**SworD** I would say that its a matter of opinion. Both are equally legit strategies for finding the limit, both make sense, both are backed by proof and by intuition.

And if one can prove L'Hospital's Rule (which is quite easy using Taylor Series), then why not use it?