Question

(13) 9. evaluate the following limits. write ∞ or -∞ for infinite limits and does not exist for limits that do not exist
a) $lim_{x
ightarrow2^{-}}\frac{x^{2}+2}{x^{2}-6x + 8}$
b) $lim_{x
ightarrow0}cos(pi/x)$
c) $lim_{x
ightarrow0}x^{4}sinleft(\frac{7pi}{x^{4}+x^{2}}
ight)$
d) $lim_{x
ightarrow0}\frac{sin(2x)}{x^{2}+4x}$
e) $lim_{x
ightarrow - 1}\frac{|x + 1|}{x + 1}$
f) $lim_{x
ightarrow0^{+}}\frac{cos x}{x}$

Explanation:

Step1: Factor the denominator for part a

Factor $x^{2}-6x + 8=(x - 2)(x - 4)$. Then $\lim_{x
ightarrow2^{-}}\frac{x^{2}+2}{x^{2}-6x + 8}=\lim_{x
ightarrow2^{-}}\frac{x^{2}+2}{(x - 2)(x - 4)}$. As $x
ightarrow2^{-}$, $x^{2}+2
ightarrow6$, $(x - 2)
ightarrow0^{-}$ and $(x - 4)
ightarrow - 2$. So the limit is $\frac{6}{0^{-}\times(-2)}=-\infty$.

Step2: Analyze part b

As $x
ightarrow0$, $\frac{\pi}{x}$ oscillates between $-\infty$ and $\infty$. The cosine - function $y = \cos t$ oscillates between $-1$ and $1$ as $t
ightarrow\pm\infty$. So $\lim_{x
ightarrow0}\cos(\frac{\pi}{x})$ does not exist.

Step3: Use the Squeeze Theorem for part c

We know that $-1\leqslant\sin(\frac{7\pi}{x^{4}+x^{2}})\leqslant1$. Then $-x^{4}\leqslant x^{4}\sin(\frac{7\pi}{x^{4}+x^{2}})\leqslant x^{4}$. Since $\lim_{x
ightarrow0}(-x^{4}) = 0$ and $\lim_{x
ightarrow0}x^{4}=0$, by the Squeeze Theorem, $\lim_{x
ightarrow0}x^{4}\sin(\frac{7\pi}{x^{4}+x^{2}})=0$.

Step4: Use the small - angle approximation for part d

We know that $\sin(2x)\sim2x$ as $x
ightarrow0$. Then $\lim_{x
ightarrow0}\frac{\sin(2x)}{x^{2}+4x}=\lim_{x
ightarrow0}\frac{2x}{x(x + 4)}=\lim_{x
ightarrow0}\frac{2}{x + 4}=\frac{1}{2}$.

Step5: Analyze the absolute - value function for part e

For $x
ightarrow - 1$, when $x\lt - 1$, $|x + 1|=-(x + 1)$, so $\lim_{x
ightarrow - 1^{-}}\frac{|x + 1|}{x + 1}=\lim_{x
ightarrow - 1^{-}}\frac{-(x + 1)}{x + 1}=-1$; when $x\gt - 1$, $|x + 1|=x + 1$, so $\lim_{x
ightarrow - 1^{+}}\frac{|x + 1|}{x + 1}=\lim_{x
ightarrow - 1^{+}}\frac{x + 1}{x + 1}=1$. Since the left - hand limit and the right - hand limit are not equal, $\lim_{x
ightarrow - 1}\frac{|x + 1|}{x + 1}$ does not exist.

Step6: Analyze part f

As $x
ightarrow0^{+}$, $\cos x
ightarrow1$ and $x
ightarrow0^{+}$. So $\lim_{x
ightarrow0^{+}}\frac{\cos x}{x}=\infty$.

Answer:

a. $-\infty$
b. does not exist
c. $0$
d. $\frac{1}{2}$
e. does not exist
f. $\infty$