Find the value of \(\,\,\,\begin{array}{*{20}{c}} {\lim }\\ {\left( {x,y} \right) \to \left( {0,0} \right)} \end{array}\frac{{121.{x^{ - 5}}.{y^{\frac{{13}}{3}}}}}{{y + {{\left( x \right)}^{\frac{3}{2}}}}}\)

Concept:

• The squeeze theorem states that if we define functions such that h(x) ≤ f(x) ≤ g(x) and if \(\rm \displaystyle \lim_{x \to a}h(x) = \lim_{x \to a}g(x) = L\), then \(\rm \displaystyle \lim_{x \to a}f(x) = L\).

Calculation:

Given:

\(\,\,\,\begin{array}{*{20}{c}} {\lim }\\ {\left( {x,y} \right) \to \left( {0,0} \right)} \end{array}\frac{{121.{x^{ - 5}}.{y^{\frac{{13}}{3}}}}}{{y + {{\left( x \right)}^{\frac{3}{2}}}}}\)

Consider approaching (0, 0) along x-axis. This means fixing y = 0 and find:

\(\mathop {\lim }\limits_{x \to 0 } ~\frac{121x^{-5}~\times~y^{\frac{13}{5}}}{y~+~x^{\frac{3}{2}}}\)

∴ \(\mathop {\lim }\limits_{{{x \to 0}}\ {y~=~0}}~\frac{121x^{-5}~\times~y^{\frac{13}{3}}}{y~+~x^{\frac{3}{2}}}~=~\frac{0}{0~+~x^{\frac{3}{2}}}~=~0\) ............................ (1)

Consider approaching (0, 0) along y-axis. This means fixing x = 0 and finding:

\(\mathop {\lim }\limits_{y \to 0 } ~\frac{121x^{-5}~\times~y^{\frac{13}{3}}}{y~+~x^{\frac{3}{2}}}\)

\(\mathop {\lim }\limits_{{{y \to 0}}\ {x~=~0}}~\frac{121x^{-5}~\times~y^{\frac{13}{3}}}{y~+~x^{\frac{3}{2}}}~=~\frac{0}{y~+~0}~=~0\) ..................................(2)

Now considering a path: y = \(x^{\frac{3}{2}}\)

\(\mathop {\lim }\limits_{{{x \to 0}}\ {y~=~x^{\frac{3}{2}}~{ \to 0}}}~\frac{121x^{-5}~\times~{(x^{\frac{3}{2}})}^{\frac{13}{3}}}{x^{\frac{3}{2}}~+~x^{\frac{3}{2}}}\)

⇒ \(\mathop {\lim }\limits_{{{x \to 0}}\ {y~=~x^{\frac{3}{2}}~{ \to 0}}}~\frac{121x^{-5~+~\frac{13}{2}}}{2~\times~x^{\frac{3}{2}}}\)

⇒ \(\mathop {\lim }\limits_{{{x \to 0}}\ {y~=~x^{\frac{3}{2}}~{ \to 0}}}~\frac{121x^{\frac{3}{2}}}{2~\times~x^{\frac{3}{2}}}~=~\frac{121}{2}\) ........................................(3)

So, From (1), (2) and (3), Limits along different paths are different, So, the limit does not exist