help with differentiation and limits

**flyinhigh123** · September 6th 2009, 04:53 AM

Differentiate y = logx3 (base is x) ie. y^3 = x

Evaluate lim xsin1/x using u=1/x
x=>∞

lim (√x - √5)/(x-5)
x=>5

thankyou so much to anyone who can help me with these questions !

**red_dog** · September 6th 2009, 07:08 AM

1) $y=\log_x3=\frac{\ln 3}{\ln x}$

and use the formula $\left(\frac{a}{y}\right)'=-\frac{ay'}{y^2}$

2) $\lim_{x\to\infty}x\sin\frac{1}{x}=\lim_{x\to\infty }\frac{\sin\frac{1}{x}}{\frac{1}{x}}$

Let $u=\frac{1}{x}$ . Then $u\to 0$ as $x\to\infty$ .

Then the limit is $\lim_{u\to 0}\frac{\sin u}{u}=1$

3) Factorize the denominator: $x-5=(\sqrt{x}-\sqrt{5})(\sqrt{x}+\sqrt{5})$

**ANDS!** · September 6th 2009, 07:10 AM

Differentiate y = logx3 (base is x) ie. y^3 = x

Well, you're going to run into problems because $y=\log[x]{3}$ does not equal $y^{3}=x$ . Remember, that the logarithim with base "a" is asking you to find the power to raise it to get "b". In this case: $x^{y}=3$ . Do you see what you do from there?

Evaluate lim xsin1/x using u=1/x
x=>∞

If $u=\frac{1}{x}$ and $x=\frac{1}{u}$ ;

Then we can rewrite this problem as $\frac{sin(u)}{u}$ . Now, we need to look at the equation $u=\frac{1}{x}$ . The limit as x approaches infinity of this equation is zero yes? Then we can say that the limit as x approaches infinity of $xsin(\frac{1}{x})$ is the same as the limit as u approaches 0 of $\frac{sin(u)}{u}$ .

Now this may not be any help if you haven't learned the squeeze theorem of haven't seen a limit of sin and x written as such.

lim (√x - √5)/(x-5)
x=>5

Oh I hated these when I first started. Because they look so dastardly, but when the instructor did it - it all seemed so easy. And it is as this can be done in a few steps:

$\frac{\sqrt{x}-\sqrt{5}}{x-5}\Rightarrow$
$\frac{\sqrt{x}-\sqrt{5}}{x-5}\\frac{\sqrt{x}+\sqrt{5}}{\sqrt{x}+\sqrt{5}}\Rig htarrow$

Can you take it from here?

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