Question

How to find the limit of square roots?

Answer 1

In the above given type of questions first we will have to decide what limits are we going to apply and also check what is the function in the square root, there are different methods of solving limit of square root, like taking common of the highest root from the denominator as well as numerator, by rationalizing either denominator or numerator, by using L’Hospital’s Rule.

Complete step-by-step solution:
Lets us take an example to this question, let the function be $y=\displaystyle \lim_{x \to 2}\dfrac{\sqrt{4x+1}-3}{x-2}$
In this we can clearly see that this limit is tending towards $\dfrac{0}{0}$ form so we are going to rationalize in terms of numerator and we will get it as, $\Rightarrow y=\left( \dfrac{\sqrt{4x+1}-3}{x-2} \right)\left( \dfrac{\sqrt{4x+1}+3}{\sqrt{4x+1}+3} \right)$
Now when we see the numerator we can see that the form that is formed is of $\left( A-B \right)\left( A+B \right)={{A}^{2}}-{{B}^{2}}$
So we can also write the numerator as

& \Rightarrow y=\dfrac{4x+1-9}{\left( x-2 \right)\left( \sqrt{4x+1}+3 \right)} \\\ & \Rightarrow y=\dfrac{4x-8}{\left( x-2 \right)\left( \sqrt{4x+1}+3 \right)} \\\ \end{aligned}$$ Now we can take 4 as common out of the numerator and we get the final equation as $$\Rightarrow y=\dfrac{4\left( x-2 \right)}{\left( x-2 \right)\left( \sqrt{4x+1}+3 \right)}$$ As we can see that x-2 is a common term in both numerator and denominator we can cancel them out which will result us in $$\Rightarrow y=\dfrac{4}{\left( \sqrt{4x+1}+3 \right)}$$ Now in this equation we can see that there is no formation of $$\dfrac{0}{0}$$ form, so we will not have to apply any other method in this example, but if after this part if we see $$\dfrac{0}{0}$$ form we will do apply the methods again and again until the whole equation is free from going towards $$\dfrac{0}{0}$$ form and as there is no formation of $$\dfrac{0}{0}$$ form we can now proceed to apply the limits that are given in the question form which we will get $$\begin{aligned} & \Rightarrow y=\displaystyle \lim_{x \to 2}\dfrac{4}{\left( \sqrt{4x+1}+3 \right)} \\\ & \Rightarrow y=\dfrac{4}{\left( \sqrt{4\left( 2 \right)+1}+3 \right)} \\\ & \Rightarrow y=\dfrac{4}{\left( \sqrt{9}+3 \right)} \\\ & \Rightarrow y=\dfrac{4}{\left( 3+3 \right)} \\\ & \Rightarrow y=\dfrac{4}{6} \\\ & \Rightarrow y=\dfrac{2}{3} \\\ \end{aligned}$$ **So after applying the limits we can clearly see that the final value of this square limit is$$y=\dfrac{2}{3}$$.** **Note:** In this type of question we need to always be careful about two things that mostly forms while applying limits, $$\dfrac{0}{0}$$ and $$\dfrac{\infty }{\infty }$$. To get rid of these and to get a final answer the above method is applied but if it is still again check what method can you apply to get rid of these forms.