Chapter 2.3, Problem 16E

To determine

To calculate: The value of the expression $\underset{y\to 2}{\lim }h\left(y\right)$ , where $h\left(y\right)=y^2\ln \left(y-1\right)$ for $y>1$ . Also estimate the value of function $0.1$ and $0.01$ above and below the limiting argument.

Answer to Problem 16E

The value of the expression $\underset{y\to 2}{\lim }h\left(y\right)$ is $0$ , the obtained limit is correct since value of function $0.1$ and $0.01$ above and below the limiting argument are $h\left(1.9\right)=-0.38$ , $h\left(2.1\right)=0.42$ , $h\left(1.99\right)=-0.04$ and $h\left(2.01\right)=0.04$ .

Explanation of Solution

Given information:

The function $h\left(y\right)=y^2\ln \left(y-1\right)$ for $y>1$ .

Formula used:

If a function f is continuous at a point say a then $\underset{x\to a}{\lim }f\left(x\right)$ is the value of the function at that point that is $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$ , plug the limit in the function to evaluate the value.

Calculation:

Consider the provided function, $h\left(y\right)=y^2\ln \left(y-1\right)$ for $y>1$ .

The value of the expression $\underset{y\to 2}{\lim }h\left(y\right)$ is evaluated when y is substitute as 2 in the expression $h\left(y\right)=y^2\ln \left(y-1\right)$ ,

  $\begin{array}{c}\underset{y\to 2}{\lim }h\left(y\right)=\underset{y\to 2}{\lim }\left[y^2\ln \left(y-1\right)\right]\\ =2^2\ln \left(2-1\right)\\ =0\end{array}$

The value of the function $h\left(y\right)=y^2\ln \left(y-1\right)$ , $0.1$ below the limiting argument that is $y=1.9$ ,

  $\begin{array}{c}h\left(1.9\right)={\left(1.9\right)}^2\ln \left(1.9-1\right)\\ =-0.38\end{array}$

The value of the function $h\left(y\right)=y^2\ln \left(y-1\right)$ , $0.1$ above the limiting argument that is $y=2.1$ ,

  $\begin{array}{c}h\left(2.1\right)={\left(2.1\right)}^2\ln \left(2.1-1\right)\\ =0.42\end{array}$

The value of the function $h\left(y\right)=y^2\ln \left(y-1\right)$ , $0.01$ below the limiting argument that is $y=1.99$ ,

  $\begin{array}{c}h\left(1.99\right)={\left(1.99\right)}^2\ln \left(1.99-1\right)\\ =-0.04\end{array}$

The value of the function $h\left(y\right)=y^2\ln \left(y-1\right)$ , $0.01$ above the limiting argument that is $y=2.01$ ,

  $\begin{array}{c}h\left(2.01\right)={\left(2.01\right)}^2\ln \left(2.01-1\right)\\ =0.04\end{array}$

Thus, the value of the expression $\underset{y\to 2}{\lim }h\left(y\right)$ is $0$ , the obtained limit is correct since value of function $0.1$ and $0.01$ above and below the limiting argument are $h\left(1.9\right)=-0.38$ , $h\left(2.1\right)=0.42$ , $h\left(1.99\right)=-0.04$ and $h\left(2.01\right)=0.04$ .