Question

Evaluate the given limit
\[\displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)\]

Answer 1

Hint: Now to evaluate the given limit we will first rationalize the numerator by multiplying and dividing the function with $\sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2}$ then we will simplify using the formula ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$ . Now again we will again rationalize the numerator by multiplying it with \[\sqrt{1+{{y}^{4}}}+1\] hence again we simplify using the formula ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$. Now we will have a limit which is not in indeterminate form. Hence we can substitute y = 0 directly to evaluate the limit

Complete step-by-step answer:
Now consider the limit \[\displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)\] if we substitute y = 0 in the limit we get $\dfrac{0}{0}$ hence this is in indeterminate form.
Let us simplify the numerator first, we want to get rid of square root hence we will try to simplify the numerator by multiplying it with $\sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2}$ . Hence we get
\[\displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}}\times \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2}}{\sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2}} \right)\]
Now we know that the formula for ${{a}^{2}}-{{b}^{2}}$ which is ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$ using this we get
\[\begin{align}
  & \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{{{\left( \sqrt{1+\sqrt{1+{{y}^{4}}}} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}{{{y}^{4}}\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right)} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{1+\sqrt{1+{{y}^{4}}}-2}{{{y}^{4}}\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right)} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+{{y}^{4}}}-1}{{{y}^{4}}\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right)} \right) \\
\end{align}\]
Now we will again try to simplify the numerator by multiplying \[\sqrt{1+{{y}^{4}}}+1\] to numerator and denominator hence we get
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+{{y}^{4}}}-1}{{{y}^{4}}\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right)}\times \dfrac{\sqrt{1+{{y}^{4}}}+1}{\sqrt{1+{{y}^{4}}}+1} \right) \\
 & \\
\end{align}\]
Now we know that the formula for ${{a}^{2}}-{{b}^{2}}$ which is ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$ using this we get
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{{{\left( \sqrt{1+{{y}^{4}}} \right)}^{2}}-{{1}^{2}}}{{{y}^{4}}\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right)\left( \sqrt{1+{{y}^{4}}}+1 \right)} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{1+{{y}^{4}}-1}{{{y}^{4}}\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right)\left( \sqrt{1+{{y}^{4}}}+1 \right)} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{{{y}^{4}}}{{{y}^{4}}\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right)\left( \sqrt{1+{{y}^{4}}}+1 \right)} \right) \\
\end{align}\]
\[\Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{1}{\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right)\left( \sqrt{1+{{y}^{4}}}-1 \right)} \right)\]
Now we can see that the limit is not in any indeterminate form. The term ${{y}^{4}}$ which made the denominator 0 is cancelled hence now we can easily substitute the value of y which is 0. To find the limits.
Hence substituting y = 0 we get
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\left( \dfrac{1}{\left( \sqrt{1+\sqrt{1+{{0}^{4}}}}+\sqrt{2} \right)\left( \sqrt{1+{{0}^{4}}}+1 \right)} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\left( \dfrac{1}{\left( \sqrt{1+\sqrt{1}}+\sqrt{2} \right)\left( \sqrt{1}+1 \right)} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}} \right)=\left( \dfrac{1}{\left( 2\sqrt{2} \right)\left( 2 \right)} \right)=\dfrac{1}{4\sqrt{2}} \\
\end{align}\]

Note: Note that while solving the limit we can use L hospital rule since the limit is in the form of $\dfrac{0}{0}$ . According to L hospital rule $\displaystyle \lim_{x \to 0}\left( \dfrac{f(x)}{g(x)} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{f'(x)}{g'(x)} \right)$. Hence we can use this to solve our limit.