MHB Evaluating a Limit: $\lim_{x\to 0}\frac{1}{4}$

[karush]

Evaluate

$$\lim_{{x}\to{0}}\frac{\sqrt{1+\tan\left({x}\right)}-\sqrt{1+\sin\left({x}\right)}}{{x}^{3}}=\frac{1}{4}$$

I tried but didn't know how to deal with the undefines

 

[MarkFL]

1.) "Rationalize" the numerator.

2.) Factor and use limit of product is product of limits.

3.) Use $$\lim_{x\to0}\frac{\sin(x)}{x}=1$$

4.) "Rationalize" numerator, then apply Pythagorean identity.

5.) Use limit of power is power of limit and $$\lim_{x\to0}\frac{\sin(x)}{x}=1$$

6.) What remains goes to 1/4 by direct substitution.

 

[karush]

by rationalizing

$$\frac{\sqrt{1+\tan\left({x}\right)}-\sqrt{1+\sin\left({x}\right)}}{{x}^{3}}=
\frac{\tan{(x)}-\sin{(x)}}{{x}^{3}\left(\sqrt{1+\tan\left({x}\right)}+\sqrt{1+\sin\left({x}\right)}\right)}$$

how could you get $\displaystyle \frac{\sin(x)}{x}$ unless you divide all terms by $x$

 

[MarkFL]

Okay, you now have:

$$\frac{\tan{(x)}-\sin{(x)}}{{x}^{3}\left(\sqrt{1+\tan\left({x}\right)}+\sqrt{1+\sin\left({x}\right)}\right)}$$

Observe that both terms in the numerator have $\sin(x)$ as a factor, so write it as:

$$\frac{\sin(x)}{x}\cdot\frac{\sec(x)-1}{{x}^{2}\left(\sqrt{1+\tan\left({x}\right)}+\sqrt{1+\sin\left({x}\right)}\right)}$$

And then continue from there, using my advice above. :D

 

[karush]

I don't see that pluging in $0$ would result in $\frac{1}{4}$

 

[MarkFL]

I don't see that pluging in $0$ would result in $\frac{1}{4}$

You still have some work to do...:D

 

[karush]

$$\lim_{{x}\to{0}}\frac{\sin\left({x}\right)}{x}
\cdot\lim_{{x}\to{0}}\frac{1}{\sqrt{1+\tan\left({x}\right)}+\sqrt{1+\sin\left({x}\right)}}
\cdot\lim_{{x}\to{0}}\frac{\sec\left({x}\right)-1}{{x}^{2 }}$$
$$1\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$$

 

[MarkFL]

This is correct:

$$\lim_{x\to0}\frac{\sec(x)-1}{x^2}=\frac{1}{2}$$

But how did you determine it?

 

[karush]

TI

But personally don't know

Thot $$\frac{du}{dv}$$ but not

 

[MarkFL]

This is what I intended:

$$\frac{\sec(x)-1}{x^2}\cdot\frac{\sec(x)+1}{\sec(x)+1}=\frac{1}{\sec(x)+1}\cdot\frac{\sec^2-1}{x^2}$$

Now, the limit of the first quotient is 1/2, and we are left with:

$$\frac{\sec^2-1}{x^2}$$

We need for this to go to 1, and so applying the Pythagorean identity $\tan^2(x)=\sec^2(x)-1$, we have:

$$\left(\frac{\tan(x)}{x}\right)^2=\sec^2(x)\left(\frac{\sin(x)}{x}\right)^2$$

And we see that this does in fact go to 1. :D

 

[karush]

I wouldn't of got to the last step
Bur that is very helpful