Question

the limit of 2x+3y/5x+4y as (x,y) --> (2,1) is​

Answer 1

Given:

A function is given to us which is (2x+3y)(5x+4y).

To Find:

We need to find the value of f(x,y) when (x,y) tend to (2,1).

Step-by-step explanation:

• We have given function we can write it as

                   $f(x,y)=\frac{2x+3y}{5x+4y}$

• Now we have value of (x,y) tends to (2,1) thus we get

             $\lim_{(x,y) \to (2,1)} f(x,y) =\frac{2x+3y}{5x+4y}$

• Put the values x=2, y=1 in above equation we get

               $\lim_{(x,y) \to (2,1)} f(x,y) =\frac{2(2)+3(1)}{5(2)+4(1)} =\frac{7}{14} =\frac{1}{2}$

Thus limit will be $\frac{1}{2}$

Answer 2

$\displaystyle\lim_{(x,y)\to (2,1)}\dfrac{2x+3y}{5x+4y}=\dfrac{1}{2}$

Step-by-step explanation:

Now, $\displaystyle\lim_{(x,y)\to (2,1)}\dfrac{2x+3y}{5x+4y}$

$\displaystyle=\dfrac{\displaystyle\lim_{(x,y)\to (2,1)}(2x+3y)}{\displaystyle\lim_{(x,y)\to (2,1)} (5x+4y)}$

$\displaystyle=\dfrac{2(2)+3(1)}{5(2)+4(1)}$

• We simply substitute the values of $x$ and $y$ in their respective places.

• Then we simplify the expression to find our required solution.

$\displaystyle=\dfrac{4+3}{10+4}$

$\displaystyle=\dfrac{7}{14}$

$\displaystyle=\dfrac{1}{2}$

Note:

Let us know a few limit formulas also that may help us solving other problems.

• $\displaystyle\lim_{x\to 0}\dfrac{sinx}{x}=1$

• $\displaystyle\lim_{x\to 0}\dfrac{tanx}{x}=1$

• $\displaystyle\lim_{x\to a}\dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}$

In the given problem we have substituted the values of $x$ and $y$ with the understanding that when $(x,y)\to (2,1)$, separately

1. $x\to 2$ and
2. $y\to 1$