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 !

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- Sep 6th 2009, 04:53 AMflyinhigh123help with differentiation and limits
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 ! - Sep 6th 2009, 07:08 AMred_dog
1) $\displaystyle y=\log_x3=\frac{\ln 3}{\ln x}$

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

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

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

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

3) Factorize the denominator: $\displaystyle x-5=(\sqrt{x}-\sqrt{5})(\sqrt{x}+\sqrt{5})$ - Sep 6th 2009, 07:10 AMANDS!Quote:

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

Quote:

Evaluate lim xsin1/x using u=1/x

x=>∞

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

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.

Quote:

lim (√x - √5)/(x-5)

x=>5

$\displaystyle \frac{\sqrt{x}-\sqrt{5}}{x-5}\Rightarrow$

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

Can you take it from here?