Question

How do you evaluate the limit ${{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}$ as $y \to 5$.

Answer 1

To find the limit here we will use direct substitution of the limit. This can be done as y = 5 into the function ${{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}$. Then we will use the following two tricks to find the cube root here. First one is, when the last digit is 4 then the cube root of a two digit number is 4 and second one is, that when the last digit is 7 so the cube root of a two digit number is 3.

Complete step by step solution:
Consider the function ${{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}$ and to find its limit as y approaches to the point 5, we will use direct substitution. As we need to find its limit nearer to y approaches to 5 so, we will put y = 5 in this function and get the limit. Therefore we get

& \displaystyle \lim_{y \to 5}{{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}={{\left( \dfrac{2{{\left( 5 \right)}^{2}}+2\left( 5 \right)+4}{6\left( 5 \right)-3} \right)}^{\dfrac{1}{3}}} \\\ & \Rightarrow \displaystyle \lim_{y \to 5}{{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}={{\left( \dfrac{50+10+4}{30-3} \right)}^{\dfrac{1}{3}}} \\\ & \Rightarrow \displaystyle \lim_{y \to 5}{{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}={{\left( \dfrac{64}{27} \right)}^{\dfrac{1}{3}}} \\\ \end{aligned}$$ Here, we will check the last digit of the numbers 64 and 27 to get their cube roots. Since, the last digit of 64 is 4 so, the cube root is simply going to be 4 only. Similarly, as the last digit of 27 is 7 so the cube root is going to be 3. By substituting 4 in place of 64 and 3 in place of 27 we get, $$\begin{aligned} & \displaystyle \lim_{y \to 5}{{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}=\dfrac{{{\left( 64 \right)}^{\dfrac{1}{3}}}}{{{\left( 27 \right)}^{\dfrac{1}{3}}}} \\\ & \Rightarrow \displaystyle \lim_{y \to 5}{{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}=\dfrac{4}{3} \\\ \end{aligned}$$ Hence, the value of the function $$\displaystyle \lim_{y \to 5}{{\left( \dfrac{2{{y}^{2}}+2y+4}{6y-3} \right)}^{\dfrac{1}{3}}}=\dfrac{4}{3}$$. **Note:** Process of factorization will not be applied here as the algebraic function in the numerator of $$\dfrac{2{{y}^{2}}+2y+4}{6y-3}$$ cannot be factorized in a simpler form. Since, after substituting y = 5 in the denominator we get no 0. Therefore, it is the right chance to substitute the value of y = 5 in the function. To find the cube rot of the number 64, one can use factorization as done below. $\begin{aligned} & 2\left| \\!{\underline {\, 64 \,}} \right. \\\ & 2\left| \\!{\underline {\, 32 \,}} \right. \\\ & 2\left| \\!{\underline {\, 16 \,}} \right. \\\ & 2\left| \\!{\underline {\, 8 \,}} \right. \\\ & 2\left| \\!{\underline {\, 4 \,}} \right. \\\ & 2\left| \\!{\underline {\, 2 \,}} \right. \\\ & \,\,\,\,1 \\\ & \Rightarrow 64=2\times 2\times 2\times 2\times 2\times 2 \\\ \end{aligned}$ After this we can combine three same numbers. As in this case the numbers are combines like so, $$\begin{aligned} & 64=\left( 2\times 2 \right)\times \left( 2\times 2 \right)\times \left( 2\times 2 \right) \\\ & \Rightarrow 64=\overline{4\times 4\times 4} \\\ & \Rightarrow {{\left( 64 \right)}^{\dfrac{1}{3}}}=4 \\\ \end{aligned}$$ Therefore, the cube root of 64 is 4. Similarly, we can find the cube root of 27 as well.