Chapter 11.2, Problem 75E

To determine

To state: which Limit is evaluated by direct substitution and find the limit

Answer to Problem 75E

Here, Limit (a) can be evaluated by direct substitution, the $\underset{x\to 0}{\lim }{x}^2\sin x^2=0$ and $\underset{x\to 0}{\lim }\frac{\sin x^2}{x^2}=1$

Explanation of Solution

Given:

  $\underset{x\to 0}{\lim }{x}^2\sin x^2$ and $\underset{x\to 0}{\lim }\frac{\sin x^2}{x^2}$

Calculation:

Limit (a) can be evaluated by using direct substitution, since it is a rational function defined at $x=0$ . Therefore :

  $\underset{x\to 0}{\lim }{x}^2\sin x^2={\left(0\right)}^2\sin {\left(0\right)}^2=0$

Limit (b) cannot be evaluated by direct substitution, since the function is undefined at $x=0$ (causes a divide by zero), so approximate this limit by using a table:

$x$	$f\left(x\right)$
$-0.1$$-0.01$$-0.001$$0$$0.001$$0.01$$0.1$	$0.999983$$1.$$1.$ Indeterminate $1.$$1.$$0.999983$

From the table, see that the function approaches the value $1$ as $x\to 0$ from both the positive and negative directions, so

  $\underset{x\to 0}{\lim }\frac{\sin x^2}{x^2}=1$

Limit (a) can be evaluated by direct substitution.

  $\underset{x\to 0}{\lim }{x}^2\sin x^2=0$

  $\underset{x\to 0}{\lim }\frac{\sin x^2}{x^2}=1$

Conclusion:

Therefore,Limit (a) can be evaluated by direct substitution, $\underset{x\to 0}{\lim }{x}^2\sin x^2=0$ and $\underset{x\to 0}{\lim }\frac{\sin x^2}{x^2}=1$