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
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
\[e^0 = 1\]
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).
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.