Answer :
The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by
[tex]\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}[/tex]
Then you want to know the value of
[tex]\displaystyle\lim_{n\to\infty}a_n[/tex]
if it exists.
To show the limit exists and that [tex]a_n[/tex] converges to some limit, we can try showing that the sequence is bounded and monotonic.
Boundedness: It's true that [tex]a_1=\sqrt{42}\le\sqrt{49}=7[/tex]. Suppose [tex]a_k\le 7[/tex]. Then [tex]a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7[/tex]. So by induction, [tex]a_n[/tex] is bounded above by 7 for all [tex]n[/tex].
Monontonicity: We have [tex]a_1=\sqrt{42}[/tex] and [tex]a_2=\sqrt{42+\sqrt{42}}[/tex]. It should be quite clear that [tex]a_2>a_1[/tex]. Suppose [tex]a_k>a_{k-1}[/tex]. Then [tex]a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k[/tex]. So by induction, [tex]a_n[/tex] is monotonically increasing.
Then because [tex]a_n[/tex] is bounded above and strictly increasing, the limit exists. Call it [tex]L[/tex]. Now,
[tex]\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n-1}=L[/tex]
[tex]\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sqrt{42+a_{n-1}}=\sqrt{42+\lim_{n\to\infty}a_{n-1}}[/tex]
[tex]\implies L=\sqrt{42+L}[/tex]
Solve for [tex]L[/tex]:
[tex]L^2=42+L\implies L^2-L-42=(L-7)(L+6)=0\implies L=7[/tex]
We omit [tex]L=-6[/tex] because our analysis above showed that [tex]L[/tex] must be positive.
So the value of the infinitely nested radical is 7.