Question

using the squeeze theorem in exercises 93 - 96, use a graphing utility to graph the given function and the equations y = |x| and y = -|x| in the same viewing window. using the graphs to observe the squeeze theorem visually, find lim(x→0)f(x). 93. f(x)=|x|sin x 94. f(x)=|x|cos x 95. f(x)=x sin(1/x) 96. f(x)=x cos(1/x)

Explanation:

Step1: Recall the range of trig - functions

We know that $- 1\leqslant\sin x\leqslant1$, $-1\leqslant\cos x\leqslant1$, and $-1\leqslant\sin\frac{1}{x}\leqslant1$, $-1\leqslant\cos\frac{1}{x}\leqslant1$.

Step2: Analyze the function $y = |x|\sin x$

Since $-1\leqslant\sin x\leqslant1$, we have $-|x|\leqslant|x|\sin x\leqslant|x|$. And $\lim_{x
ightarrow0}-|x| = 0$ and $\lim_{x
ightarrow0}|x| = 0$. By the Squeeze Theorem, $\lim_{x
ightarrow0}|x|\sin x=0$.

Step3: Analyze the function $y = |x|\cos x$

Since $-1\leqslant\cos x\leqslant1$, we have $-|x|\leqslant|x|\cos x\leqslant|x|$. And $\lim_{x
ightarrow0}-|x| = 0$ and $\lim_{x
ightarrow0}|x| = 0$. By the Squeeze Theorem, $\lim_{x
ightarrow0}|x|\cos x=0$.

Step4: Analyze the function $y = x\sin\frac{1}{x}$

Since $-1\leqslant\sin\frac{1}{x}\leqslant1$, we have $-|x|\leqslant x\sin\frac{1}{x}\leqslant|x|$ (when considering the absolute - value inequality). And $\lim_{x
ightarrow0}-|x| = 0$ and $\lim_{x
ightarrow0}|x| = 0$. By the Squeeze Theorem, $\lim_{x
ightarrow0}x\sin\frac{1}{x}=0$.

Step5: Analyze the function $y = x\cos\frac{1}{x}$

Since $-1\leqslant\cos\frac{1}{x}\leqslant1$, we have $-|x|\leqslant x\cos\frac{1}{x}\leqslant|x|$. And $\lim_{x
ightarrow0}-|x| = 0$ and $\lim_{x
ightarrow0}|x| = 0$. By the Squeeze Theorem, $\lim_{x
ightarrow0}x\cos\frac{1}{x}=0$.

Answer:

For $f(x)=|x|\sin x$, $\lim_{x
ightarrow0}f(x) = 0$; for $f(x)=|x|\cos x$, $\lim_{x
ightarrow0}f(x)=0$; for $f(x)=x\sin\frac{1}{x}$, $\lim_{x
ightarrow0}f(x)=0$; for $f(x)=x\cos\frac{1}{x}$, $\lim_{x
ightarrow0}f(x)=0$