find the limit.\n61) $lim_{x\rightarrow(-2)^-}\frac{1}{x + 2}$\n62) $lim_{x\rightarrow(-2)^+}\frac{1}{x +…

find the limit.\n61) $lim_{x\rightarrow(-2)^-}\frac{1}{x + 2}$\n62) $lim_{x\rightarrow(-2)^+}\frac{1}{x + 2}$\n63) $lim_{x\rightarrow - 2}\frac{1}{x + 2}$\n64) $lim_{x\rightarrow0}\frac{1}{x^{2/3}}$\n65) $lim_{x\rightarrow(pi/2)^+}\tan x$\n66) $lim_{x\rightarrow(-pi/2)^-}sec x$\n67) $lim_{x\rightarrow0^+}(1+csc x)$\n68) $lim_{x\rightarrow0}(1 - cot x)$

find the limit.\n61) $lim_{x\rightarrow(-2)^-}\frac{1}{x + 2}$\n62) $lim_{x\rightarrow(-2)^+}\frac{1}{x + 2}$\n63) $lim_{x\rightarrow - 2}\frac{1}{x + 2}$\n64) $lim_{x\rightarrow0}\frac{1}{x^{2/3}}$\n65) $lim_{x\rightarrow(pi/2)^+}\tan x$\n66) $lim_{x\rightarrow(-pi/2)^-}sec x$\n67) $lim_{x\rightarrow0^+}(1+csc x)$\n68) $lim_{x\rightarrow0}(1 - cot x)$

Answer

Explanation:

61)

Step1: Analyze left - hand limit

As (x\to(- 2)^{-}), let (x=-2 - h) where (h\to0^{+}). Then (\frac{1}{x + 2}=\frac{1}{-2 - h+2}=-\frac{1}{h}). As (h\to0^{+}), (\lim_{x\to(-2)^{-}}\frac{1}{x + 2}=-\infty).

62)

Step1: Analyze right - hand limit

As (x\to(-2)^{+}), let (x=-2 + h) where (h\to0^{+}). Then (\frac{1}{x + 2}=\frac{1}{-2 + h+2}=\frac{1}{h}). As (h\to0^{+}), (\lim_{x\to(-2)^{+}}\frac{1}{x + 2}=\infty).

63)

Step1: Check one - sided limits

Since (\lim_{x\to(-2)^{-}}\frac{1}{x + 2}=-\infty) and (\lim_{x\to(-2)^{+}}\frac{1}{x + 2}=\infty), the two - sided limit (\lim_{x\to - 2}\frac{1}{x + 2}) does not exist.

64)

Step1: Analyze limit as (x\to0)

As (x\to0), if (x\to0^{+}), (\frac{1}{x^{2/3}}=\frac{1}{\sqrt[3]{x^{2}}}\to+\infty), and if (x\to0^{-}), (\frac{1}{x^{2/3}}=\frac{1}{\sqrt[3]{x^{2}}}\to+\infty). So (\lim_{x\to0}\frac{1}{x^{2/3}}=\infty).

65)

Step1: Recall tangent function

(\tan x=\frac{\sin x}{\cos x}). As (x\to(\frac{\pi}{2})^{+}), (\cos x\to0^{-}) and (\sin x\to1). So (\lim_{x\to(\frac{\pi}{2})^{+}}\tan x=-\infty).

66)

Step1: Recall secant function

(\sec x=\frac{1}{\cos x}). As (x\to(-\frac{\pi}{2})^{-}), (\cos x\to0^{-}). So (\lim_{x\to(-\frac{\pi}{2})^{-}}\sec x=-\infty).

67)

Step1: Recall cosecant function

(\csc x=\frac{1}{\sin x}). As (x\to0^{+}), (\sin x\to0^{+}), so (\csc x\to+\infty). Then (\lim_{x\to0^{+}}(1 + \csc x)=\infty).

68)

Step1: Recall cotangent function

(\cot x=\frac{\cos x}{\sin x}). As (x\to0), (\lim_{x\to0}\cos x = 1) and (\lim_{x\to0}\sin x=0). (\lim_{x\to0}\cot x=\infty), so (\lim_{x\to0}(1-\cot x)=-\infty).

Answer:

  1. (-\infty)
  2. (\infty)
  3. Does not exist
  4. (\infty)
  5. (-\infty)
  6. (-\infty)
  7. (\infty)
  8. (-\infty)