Help me solve this function limit puzzle

[jraks]

hi guys got a little query

is limit of a function consisting of two variables defined when both the variables are tending to a value simultaneously
eg: lim (x/y)
x->infinity
y->0
just give it atry and if this is wrong just justify
thanks
P.S tis ones for you neils_arm_strong

 

[elmolla]

Re: limit puzzle

Simple
∞

 

[jraks]

limit puzzle

isnt infinity/0 undefined
think anout it

 

[goumaiy]

Re: limit puzzle

Yes, lim (x/y)=∞. If this is not clear, just switch the denominator and the numerator. You get lim (y/x)=0. Thus, lim (x/y)=∞.

This (∞/0, 0/∞) is perfectly defined. ∞/∞ and 0/0 are undefined.

 

[mobile]

limit puzzle

∞ is correct answer?

 

[kishangk]

Re: limit puzzle

very easy...infinity....

 

[kar2kn]

Re: limit puzzle

answer is 0.
the rule is according to L'Hospital .

 

[jayc]

Re: limit puzzle

L'hospital's does not apply here. The rule is used only if the limit goes to
\[\frac{0}{0}\] or \[\frac{\infty}{\infty}\].

 

[sp002]

Re: limit puzzle

∞ is correct answer.

 

[jraks]

limit puzzle

ok guys how about

lim x^y
x->infinity
y->0

solve it

and on more thing is limit defined at apoint when the two variables are simulataneously varying and tending to some thing?

think

 

[jayc]

Re: limit puzzle

To solve this, we first need to express the limit in a different form.

Consider the following:
\[\lim\limit_{x\rightarrow\infty, y\rightarrow 0} x^y = \lim\limit_{x\rightarrow\infty, y\rightarrow 0} e^{\ln(x^y)}\]

Let us first find the limit of the exponent:
\[\lim\limit_{x\rightarrow\infty, y\rightarrow 0} \ln x^y = \lim\limit_{x\rightarrow\infty, y\rightarrow 0} y \ln x\]

This equates to 0*inf so we need to apply L'Hospital's Rule:
\[\lim\limit_{x\rightarrow\infty, y\rightarrow 0} \frac{\ln x}{1/y}\]
\[{} = \lim\limit_{x\rightarrow\infty, y\rightarrow 0} \frac{1/x}{-1/y^2} = \lim\limit_{x\rightarrow\infty, y\rightarrow 0} -\frac{y^2}{x}\]

It is now apparent that this limit goes to 0/inf, which is just 0.

Now we can go back to the original limit of interest by raising the result in the base e. This gives the result
\[e^0 = 1\]

 

[LouisSheffield]

Re: limit puzzle

Let's not forget, too, that it hasn't been specified as to whether y->0 from the left OR from the right.
Also, we can't apply L'Hopital's rule unless we have a functional form that relates x and y. For example, if y=1/x, then x/y=x^2, which is always positive (etc).

 

[jayc]

Re: limit puzzle

Let's not forget, too, that it hasn't been specified as to whether y->0 from the left OR from the right.
Also, we can't apply L'Hopital's rule unless we have a functional form that relates x and y. For example, if y=1/x, then x/y=x^2, which is always positive (etc).

In this problem, it doesn't matter from which direction y approaches the limit, since there is no divergence at the limit case. Also, we do not need to know a relation between x and y. This is because the numerator and denominator only contain one term and are completely independent. L'Hospital's Rule is useful because it determines the limit by finding which term (numerator or denominator) changes faster. Since they are independent, there is no problem with taking the derivative of the numerator with respect to x and the denominator with respect to y.

 

[LouisSheffield]

Re: limit puzzle

But there is divergence - x is clearly positive, but not y.
At least y was not stated as being positive as it approaches 0.

As for L'Hopital, one must have a function in order to differentiate said function.

Also, the function you specified above (x^y) is not what the original question involved. (the example was x/y, which is divergent)


Just noticed - Sorry about any confusion - my post was to the original problem x/y,
not the solution you formulated a few posts ago.