L'Hopital's Rule, 2 Confusing ones

[verd]

Hey,

So I'm having a bit of difficulty with two of these L'Hospital's Rule problems... The first:
$$\mathop {\lim }\limits_{x \to \infty} (\sqrt{x^2 + x} - x)$$

So when you have an indefinite form $$\infty - \infty$$, you've got to turn it into a product indefinite form, usually something like $$\infty * 0$$

So I take that,
$$\mathop {\lim }\limits_{x \to \infty} (\sqrt{x^2 + x} - x) = \mathop {\lim }\limits_{x \to \infty} x(\frac{\sqrt{x^2 +x}}{x} - 1)$$

From there, I try to take the limit of the fraction in the parenthesis, but end up going in circles with it. After I differentiate it the first time and get another indefinite form, I differentiate again and from there, it loops. I keep getting the reciprocal of what I started with. I know that the answer is supposed to come out to be 1/2, but I have no idea how to get there. Any ideas?The second one I'm having difficulty with is,
$$\mathop {\lim }\limits_{x \to 1+} lnx tan(\pi x/2)$$

I'm getting a similar problem with this one. No matter which way I go when I try to turn this one into a fraction, I either get the wrong answer (-1), or it just gets continually more complex. ...The answer here is supposed to be $$-2/\pi$$

Any pointers?

 

[quasar987]

Hey,

So when you have an indefinite form $$\infty - \infty$$, you've got to turn it into a product indefinite form, usually something like $$\infty * 0$$

And after that, you got to turn it into $\infty / \infty$ or $0 / 0$, because that's when l'Hospital applies.

Useful hint:

$$x = \frac{1}{\frac{1}{x}}$$

 

[TD]

The second one I'm having difficulty with is,
$$\mathop {\lim }\limits_{x \to 1+} lnx tan(\pi x/2)$$

I'm getting a similar problem with this one. No matter which way I go when I try to turn this one into a fraction, I either get the wrong answer (-1), or it just gets continually more complex. ...The answer here is supposed to be $$-2/\pi$$

Again, you get the indeterminate form $0 \cdot \infty$ but you cannot apply L'Hospital then. As quasar987 pointed out, you need something of the form 0/0 or ∞/∞: use his hint to obtain such a form.

 

[dextercioby]

Useful hint

$$\lim_{x\rightarrow +\infty} \left(\sqrt{x^{2}+x}-x\right)=\lim_{x\rightarrow +\infty} \left(\left(\sqrt{x^{2}+x}-x\right) \cdot \frac{\sqrt{x^{2}+x}+x}{\sqrt{x^{2}+x}+x}\right)$$

Daniel.

 

[dextercioby]

For the second

$$\lim_{x\rightarrow 1^{+}}\ln x\cdot \tan\frac{\pi x}{2}$$

can be solved easily if you do these steps

1. Make the substitution $x=1+y$. It follows that the new limit will be

$$\lim_{y\rightarrow 0^{+}} \left(\mbox{something}\right)$$

2. Use the approximation (coming from the MacLaurin expansion of the natural logarithm)

$$\ln\left(1+y)\simeq y$$, valid when "y" is very close to 0.

3.Use the definition of tangent wrt to sine & cosine.

4.Use the fact that

$$\lim_{y\rightarrow 0^{+}} \sin {}\frac{\pi}{2}\left(y+1\right) = 1$$.

5. Use l'Hôpital's rule.

Final result, of course $-\frac{2}{\pi}$.


Daniel.