Consider the provide expression,

\(\lim_{x\rightarrow4}\frac{31x-42\cdot\sqrt{x+5}}{3-\sqrt{x+5}}\)

Now, find the limit of the provided expression.

\(\lim_{x\rightarrow4}\frac{31x-42\cdot\sqrt{x+5}}{3-\sqrt{x+5}}\)

Its is not determinant form because the expression from is \(\frac{0}{0}\)

Apply the L Hospital rule,

\(\lim_{x\rightarrow4}\frac{31x-42\cdot\sqrt{x+5}}{(3-\sqrt{x+5}}=\lim_{x\rightarrow4}\frac{31-42}{(2\sqrt{x+5})}(3-\frac{1}{\sqrt{x+5}})\)

So, it is a determinant form

\(\lim_{x\rightarrow4}\frac{31x-42\cdot\sqrt{x+5}}{3-\sqrt{x+5}}=\lim_{x\rightarrow4}\frac{31-\frac{42}{2\cdot\sqrt{x+5}}}{3-\frac{1}{\sqrt{x+5}}}\)

\(=\frac{31-\frac{42}{6}}{3-\frac{1}{3}}\)

\(=\frac{24\cdot3}{8}\)

\(=9\)

Hence, the limit is 9

\(\lim_{x\rightarrow4}\frac{31x-42\cdot\sqrt{x+5}}{3-\sqrt{x+5}}\)

Now, find the limit of the provided expression.

\(\lim_{x\rightarrow4}\frac{31x-42\cdot\sqrt{x+5}}{3-\sqrt{x+5}}\)

Its is not determinant form because the expression from is \(\frac{0}{0}\)

Apply the L Hospital rule,

\(\lim_{x\rightarrow4}\frac{31x-42\cdot\sqrt{x+5}}{(3-\sqrt{x+5}}=\lim_{x\rightarrow4}\frac{31-42}{(2\sqrt{x+5})}(3-\frac{1}{\sqrt{x+5}})\)

So, it is a determinant form

\(\lim_{x\rightarrow4}\frac{31x-42\cdot\sqrt{x+5}}{3-\sqrt{x+5}}=\lim_{x\rightarrow4}\frac{31-\frac{42}{2\cdot\sqrt{x+5}}}{3-\frac{1}{\sqrt{x+5}}}\)

\(=\frac{31-\frac{42}{6}}{3-\frac{1}{3}}\)

\(=\frac{24\cdot3}{8}\)

\(=9\)

Hence, the limit is 9