MA005: Calculus I

Sometimes it is difficult to work directly with a function. However, if we can compare our difficult function with easier ones, then we can use information about the easier functions to draw conclusions about the difficult one. If the complicated function is always between two functions whose limits are equal, then we know the limit of the complicated function. 

Example 4: Use the inequality $-|\mathrm{x}| \leq \sin (\mathrm{x}) \leq|\mathrm{x}|$ to determine $\lim\limits_{x \rightarrow 0} \sin (\mathrm{x})$ and $\lim\limits_{x \rightarrow 0} \cos (x)$.

Solution: $\lim\limits_{x \rightarrow 0}|\mathrm{x}|=0$ and $\lim\limits_{x \rightarrow 0}-|\mathrm{x}|=0$ so, by the Squeezing Theorem,

$\lim\limits_{x \rightarrow 0} \sin (x)=0$. If $-\pi / 2$ then $\cos (x)=+\sqrt{1-\sin ^{2}(x)}$ so

$\lim\limits_{x \rightarrow 0} \cos (\mathrm{x})=\lim\limits_{x \rightarrow 0}+\sqrt{1-\sin ^{2}(\mathrm{x})}=+\sqrt{1-0}=1$

Example 5: Evaluate $\lim\limits_{x \rightarrow 0} \mathrm{x} \cdot \sin \left(\frac{1}{\mathrm{X}}\right)$.

Solution: The graph of $\mathrm{y}=\sin \left(\frac{1}{\mathrm{x}}\right)$ for values of $\mathrm{x}$ near 0 is shown in Fig. 8. The y-values of this graph change very rapidly for values of $x$ near 0, but they all lie between $-1$ and $+1$:

$-1 \leq \sin \left(\frac{1}{\mathrm{x}}\right) \leq+1.$ The fact that $\sin \left(\frac{1}{\mathrm{x}}\right)$ is bounded between $-1$ and $+1$ implies that $\mathrm{x} \sin \left(\frac{1}{\mathrm{x}}\right)$ is stuck between $-\mathrm{x}$ and $+\mathrm{x}$, so the function we are interested in, $\mathrm{x} \cdot \sin \left(\frac{1}{\mathrm{x}}\right)$, is squeezed between two "easy" functions, $-\mathrm{x}$ and $\mathrm{x}$

(Fig. 9). Both "easy" functions approach 0 as $x \rightarrow 0$, so $x \cdot \sin \left(\frac{1}{x}\right)$ must also approach 0 as $\mathrm{x} \rightarrow 0: \lim\limits_{x \rightarrow 0} \mathrm{x} \cdot \sin \left(\frac{1}{\mathrm{x}}\right)=0$.

Practice 5: If $\mathrm{f}(\mathrm{x})$ is always between $\mathrm{x}^{2}+2$ and $2 \mathrm{x}+1$, then $\lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x})=$?

Practice 6: Use the relation $\cos (\mathrm{x}) \leq \frac{\sin (\mathrm{x})}{\mathrm{x}} \leq 1$ to show that $\lim\limits_{x \rightarrow 0} \frac{\sin (x)}{x}=1$. (The steps for deriving the inequalities are shown in problem 19).