MHB Finding Limits Using Limit Laws (Not L'Hopital's Rule)

[ardentmed]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For a, I used the subtraction limit lawto get lim g(x) and lim f(x) and subtracted the answers accordingly. Then I substituted h= infinity and got 0.

For b, I applied the subtraction limit law and split the limit into two, and assessed for the limit. This gave me:
√5 / 0 - √ 5 / 0 = infinity (undefined).

For c, I split up the limit again and took the limits individually; one with the radical and the other as x by itself. This gave me infinity - infinity = 0.

As for d, I got -1 and 1 due to the absolute value, which are not equal. Thus, the limit does not exist.

Finally, for e, knowing that:

$\lim_{{x}\to{0}} $ 2π / x = infinity, I applied this to the function sin(x).

This gave me:
$\lim_{{x}\to{0}} $ sin(infinity) = 1

Taking the limit gave me √x * 1 = 0.

Any help would be greatly appreciated.

Thanks in advance.

 

[evinda]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:For a, I used the subtraction limit lawto get lim g(x) and lim f(x) and subtracted the answers accordingly. Then I substituted h= infinity and got 0.

For b, I applied the subtraction limit law and split the limit into two, and assessed for the limit. This gave me:
√5 / 0 - √ 5 / 0 = infinity (undefined).

For c, I split up the limit again and took the limits individually; one with the radical and the other as x by itself. This gave me infinity - infinity = 0.

As for d, I got -1 and 1 due to the absolute value, which are not equal. Thus, the limit does not exist.

Finally, for e, knowing that:

$\lim_{{x}\to{0}} $ 2π / x = infinity, I applied this to the function sin(x).

This gave me:
$\lim_{{x}\to{0}} $ sin(infinity) = 1

Taking the limit gave me √x * 1 = 0.

Any help would be greatly appreciated.

Thanks in advance.

$$\lim_{h \to 0} \frac{(8+h)^{-1}-8^{-1}}{h}=\lim_{h \to 0} \frac{\frac{1}{8+h}-\frac{1}{8}}{h}=\lim_{h \to 0} \frac{\frac{8-(8+h)}{8(8+h)}}{h}=\lim_{h \to 0} \frac{\frac{-h}{8(8+h)}}{h}=\lim_{h \to 0} \frac{-1}{8(8+h)}=-\frac{1}{64}$$

For the second one:

$$\frac{\sqrt{x^2+1}-\sqrt{5}}{x-2}=\frac{(\sqrt{x^2+1}- \sqrt{5})(\sqrt{x^2+1}+ \sqrt{5})}{(x-2)(\sqrt{x^2+1}+ \sqrt{5})}=\frac{x^2-4}{(x-2)(\sqrt{x^2+1}+ \sqrt{5})}=\frac{x+2}{\sqrt{x^2+1}+ \sqrt{5}}$$

Now take the limit $x \to 2$ and you will see that it is not undefined.

For c:

$$\sqrt{x^2+x+1}-x=\frac{(\sqrt{x^2+x+1}-x)(\sqrt{x^2+x+1}+x)}{\sqrt{x^2+x+1}+x}=\frac{x+1}{\sqrt{x^2+x+1}+x}=\frac{x+1}{x \left ( \sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1 \right )}$$

Now take the limit $x \to +\infty$.

You are right that the limit of the question d does not exist.

For e,notice that the limit $\lim_{x \to 0} \sin(\frac{2 \pi}{x})$ does not exist.

 

[Dethrone]

I'm not sure if you are familiar with derivatives, but I want to point out a unique solution to a) and b) These are in a form that is awful similar to the definition of a derivative.
$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
$$\lim_{x \to a} \frac{f(x)-f(a)}{x-a}
$$a) $$\lim_{h \to 0} \frac{\frac{1}{8+h}-\frac{1}{8}}{h}$$

We can rewrite this as: $$\lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$ at x = 8 = $$f'(8)$$ = $$\frac{-1}{64}$$

b) $$\lim_{x \to 2} \frac{\sqrt{x^2+1}-\sqrt{5}}{x-2}$$

We can rewrite this as: $$\lim_{x \to 2} \frac{f(x)-f(a)}{x-a}$$, where x = 2
$$= f'(2) = \frac{2}{\sqrt{5}}$$

 

[ardentmed]

$$\lim_{h \to 0} \frac{(8+h)^{-1}-8^{-1}}{h}=\lim_{h \to 0} \frac{\frac{1}{8+h}-\frac{1}{8}}{h}=\lim_{h \to 0} \frac{\frac{8-(8+h)}{8(8+h)}}{h}=\lim_{h \to 0} \frac{\frac{-h}{8(8+h)}}{h}=\lim_{h \to 0} \frac{-1}{8(8+h)}=-\frac{1}{64}$$

For the second one:

$$\frac{\sqrt{x^2+1}-\sqrt{5}}{x-2}=\frac{(\sqrt{x^2+1}- \sqrt{5})(\sqrt{x^2+1}+ \sqrt{5})}{(x-2)(\sqrt{x^2+1}+ \sqrt{5})}=\frac{x^2-4}{(x-2)(\sqrt{x^2+1}+ \sqrt{5})}=\frac{x+2}{\sqrt{x^2+1}+ \sqrt{5}}$$

Now take the limit $x \to 2$ and you will see that it is not undefined.

For c:

$$\sqrt{x^2+x+1}-x=\frac{(\sqrt{x^2+x+1}-x)(\sqrt{x^2+x+1}+x)}{\sqrt{x^2+x+1}+x}=\frac{x+1}{\sqrt{x^2+x+1}+x}=\frac{x+1}{x \left ( \sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1 \right )}$$

Now take the limit $x \to +\infty$.

You are right that the limit of the question d does not exist.

For e,notice that the limit $\lim_{x \to 0} \sin(\frac{2 \pi}{x})$ does not exist.

For c, I computed 1/2, albeit I'm not too sure about that. I canceled out the x's first, and then substituted x-> infinity for the rest.

As for e, $\lim_{x \to 0} \sin(\frac{2 \pi}{x})$ due to squeeze theorem, where -1/x and 1/x also DNE.

 

[evinda]

For c, I computed 1/2, albeit I'm not too sure about that. I canceled out the x's first, and then substituted x-> infinity for the rest.

$$\lim_{x \to +\infty} \frac{x+1}{x( \sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1)}=\lim_{x \to \infty} \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1}=\frac{1}{2}$$

So,your result is correct! (Clapping)

 

[evinda]

As for e, $\lim_{x \to 0} \sin(\frac{2 \pi}{x})$ due to squeeze theorem, where -1/x and 1/x also DNE.

For e,as you correctly mentioned,we know that:

$$-1 \leq \sin \left ( \frac{2 \pi}{x}\right ) \leq 1$$

So:
$$-\sqrt{x} \leq \sqrt{x} \sin \left ( \frac{2 \pi}{x}\right ) \leq \sqrt{x} \Rightarrow \lim_{x \to 0} -\sqrt{x} \leq \lim_{x \to 0} \sqrt{x} \sin \left ( \frac{2 \pi}{x}\right ) \leq \lim_{x \to 0} \sqrt{x} \Rightarrow 0 \leq \lim_{x \to 0} \sqrt{x} \sin \left ( \frac{2 \pi}{x}\right ) \leq 0$$

Therefore,according to the squeeze theorem:

$$\lim_{x \to 0} \sqrt{x} \sin \left ( \frac{2 \pi}{x} \right )=0$$