Find the limit.
-$$\lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { 10 x ^ { 2 } + 4 y ^ { 2 } + 1 } { 10 x ^ { 2 } - 4 y ^ { 2 } + 4 }$$
A) 1
B) $- 1$
C) $\frac { 1 } { 4 }$
D) No limit

Question

Quizplus · Accepted Answer

We can simplify the expression by dividing the numerator and denominator by $10x^2$. This gives us:
$$\lim_{(x,y)ightarrow(0,0)}\frac{10x^2+4y^2+1}{10x^2-4y^2+4}=\lim_{(x,y)ightarrow(0,0)}\frac{1+\frac{4}{10}\left(\frac{y}{x}ight)^2}{1-\frac{4}{10}\left(\frac{y}{x}ight)^2+\frac{4}{10x^2}}.$$
As $(x,y)ightarrow(0,0)$, we have $\frac{y}{x}ightarrow0$, so the limit becomes:
$$\lim_{\frac{y}{x}ightarrow 0}\frac{1+\frac{4}{10}\left(\frac{y}{x}ight)^2}{1-\frac{4}{10}\left(\frac{y}{x}ight)^2+\frac{4}{10x^2}}.$$
If we approach along the line $y=0$, then the limit is simply:
$$\lim_{\frac{y}{x}ightarrow 0}\frac{1+\frac{4}{10}\left(\frac{y}{x}ight)^2}{1-\frac{4}{10}\left(\frac{y}{x}ight)^2+\frac{4}{10x^2}}=\lim_{xightarrow 0}\frac{1}{1+\frac{4}{10x^2}}=\frac{1}{1}=\boxed{	extbf{(A) }1}.$$
If we approach along the line $y=mx$, where $m$ is some constant, then the limit is:
\begin{align*}
\lim_{\frac{y}{x}ightarrow m}\frac{1+\frac{4}{10}\left(\frac{y}{x}ight)^2}{1-\frac{4}{10}\left(\frac{y}{x}ight)^2+\frac{4}{10x^2}}&=\frac{1+\frac{4}{10}m^2}{1-\frac{4}{10}m^2}.
\end{align*}
This limit depends on the value of $m$, so the limit as $(x,y)ightarrow(0,0)$ does not exist. Therefore, the answer is $\boxed{	extbf{(C) }\frac{1}{4}}$.

Find the Limit lim⁡(x,y)→(0,0)10x2+4y2+110x2−4y2+4\lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { 10 x ^ { 2 } + 4 y ^ { 2 } + 1 } { 10 x ^ { 2 } - 4 y ^ { 2 } + 4 }(x,y)→(0,0)lim​10x2−4y2+410x2+4y2+1​

Find the Limit $\lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { 10 x ^ { 2 } + 4 y ^ { 2 } + 1 } { 10 x ^ { 2 } - 4 y ^ { 2 } + 4 }$