Q. 27 - Calculus

Question

Calculate each of the limits in Exercises 21–48. Some of these limits are made easier by L’Hopital’s rule, and some are not 

$\underset{x\to 0}{\lim }\frac{x^{-2}}{x^{-3}+1}$

Step-by-Step Solution

Answer

The value of the given limit $\underset{x\to 0}{\lim }\frac{x^{-2}}{x^{-3}+1}=0$

Step 1. Given information

The given limit$\underset{x\to 0}{\lim }\frac{x^{-2}}{x^{-3}+1}$

Step 2. Using L’Hopital’s rule find the value of given limit

The limit using L’Hopitals rule is given below:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{x^{-2}}{x^{-3}+1}=\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^3}}+1} \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1+x^2}{x^3}}} \\ =\underset{x\to 0}{\lim }\frac{1}{x^2}.\frac{x^3}{1+x^3} \\ =\underset{x\to 0}{\lim }\frac{x}{1+x^2} \\ =\frac{0}{1+0} \\ =0 \end{gathered}$$

Therefore,the value of the given limit $=0$