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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- September 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 ! - September 6th 2009, 07:08 AMred_dog
1)

and use the formula

2)

Let . Then as .

Then the limit is

3) Factorize the denominator: - September 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 . Now, we need to look at the equation . The limit as x approaches infinity of this equation is zero yes? Then we can say that the limit as x approaches infinity of is the same as the limit as u approaches 0 of .

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

Can you take it from here?