How do you use continuity to evaluate the limit $\dfrac{{{e}^{{{x}^{2}}}}-{{e}^{-{{y}^{2}}}}}{x+y}$ as $\left( x,y \right)$ approaching to $\left( 1,1 \right)$?

Hint: To evaluate the given limit $\dfrac{{{e}^{{{x}^{2}}}}-{{e}^{-{{y}^{2}}}}}{x+y}$ as $\left( x,y \right)$ approaching to $\left( 1,1 \right)$ using continuity, we are going to substitute $\left( x,y \right)$ as $\left( 1,1 \right)$ in $\dfrac{{{e}^{{{x}^{2}}}}-{{e}^{-{{y}^{2}}}}}{x+y}$ and then solve this function in (x, y) to get the value of this limit.

Complete step by step answer:
The limit given in the above problem which we have to evaluate using continuity is as follows:
$\dfrac{{{e}^{{{x}^{2}}}}-{{e}^{-{{y}^{2}}}}}{x+y}$
It is also given that $\left( x,y \right)$ is approaching $\left( 1,1 \right)$ in the above function. So, applying this liit in the above function we get,
$\displaystyle \lim_{\left( x,y \right) \to \left( 1,1 \right)}\dfrac{{{e}^{{{x}^{2}}}}-{{e}^{-{{y}^{2}}}}}{x+y}$
On using the continuity for above limit evaluation we are going to substitute the value of $\left( x,y \right)$ as $\left( 1,1 \right)$ and we get,
$\begin{align}
  & \dfrac{{{e}^{{{\left( 1 \right)}^{2}}}}-{{e}^{-{{\left( 1 \right)}^{2}}}}}{1+1} \\
 & \Rightarrow \dfrac{e-{{e}^{-1}}}{2} \\
 & \\
\end{align}$
$\begin{align}
  & \Rightarrow \dfrac{e-\dfrac{1}{e}}{2} \\
 & \Rightarrow \dfrac{{{e}^{2}}-1}{2e} \\
\end{align}$
We know that the value of “e” is equal to 2.7 so substituting “e” as 2.7 in the above we get,
$\begin{align}
  & \dfrac{{{\left( 2.7 \right)}^{2}}-1}{2\left( 2.7 \right)} \\
 & \Rightarrow \dfrac{7.29-1}{5.4} \Rightarrow \dfrac{6.29}{5.4}\approx 1.16 \\
\end{align}$
From the above, we have evaluated the limit using continuity when $\left( x,y \right)$ approaches to $\left( 1,1 \right)$ as $\left( \approx 1.16 \right)$.

Note:
The mistake that could happen in solving the above problem is that you might forget to write negative signs in the second exponential after putting x and y values.
$\dfrac{{{e}^{{{\left( 1 \right)}^{2}}}}-{{e}^{{{\left( 1 \right)}^{2}}}}}{1+1}$
Now, in the above, as you can see that there is no negative sign in the power of exponential so this is the mistake that we have talked about in the above so make sure you won’t commit such a mistake in the examination.
Also, sometimes, the examiner knows that such mistakes could be possible so the examiner intentionally put the wrong option. Like, in the above, we have shown the mistake that could be possible so if we do this mistake then we get the evaluation of limit as 0 so the wrong option that the examiner can put is 0.