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 !
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: $\displaystyle x-5=(\sqrt{x}-\sqrt{5})(\sqrt{x}+\sqrt{5})$
Well, you're going to run into problems because $\displaystyle y=\log[x]{3}$ does not equal $\displaystyle y^{3}=x$. Remember, that the logarithim with base "a" is asking you to find the power to raise it to get "b". In this case: $\displaystyle x^{y}=3$. Do you see what you do from there?Differentiate y = logx3 (base is x) ie. y^3 = x
If $\displaystyle u=\frac{1}{x}$ and $\displaystyle x=\frac{1}{u}$;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.
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: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?