Limits – Problems
I. Solving the following limits
1) \(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{2}}-{{a}^{2}}}{x-a}\)
Solution: Given that
\(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{2}}-{{a}^{2}}}{x-a}\),
\(\underset{x\to a}{\mathop{\lim }}\,\frac{(x-a)(x+a)}{x-a}\),
\(\underset{x\to a}{\mathop{\lim }}\,(x+a)\),
a + a = 2a
2. \(\underset{x\to 1}{\mathop{\lim }}\,({{x}^{2}}+2x+3)\)
Solution: Given \(\underset{x\to 1}{\mathop{\lim }}\,({{x}^{2}}+2x+3)\)
1² + 2 (1) + 3 = 6
3. \(\underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}.\cos \left( \frac{2}{x} \right)\)
Solution: Given
\(\underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}.\cos (\frac{2}{x})\)= 0. k where |K| ≤ 1 = 0
II) Find the right and left-hand limits of the function
\(f(x)=\left\{ \begin{align} & 1-x\,\ \ \ if\ x\le 1 \\ & 1+x\ \ \ \ if\ x>1 \\\end{align} \right.\ ;a=1\)Solution: Given
\(f(x)=\left\{ \begin{align} & 1-x\ \ \ \,if\ x\le 1 \\ & 1+x\ \ \ \ if\ x>1 \\\end{align} \right.\ ;a=1\)Left limit at x = 1 is \(\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,(1-x)=1-1=0\)
Right limit at x = 1 is \(\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,(1+x)=1+1=2\)
\(\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\)Therefore \(\underset{x\to 1}{\mathop{\lim }}\,f(x)\) does not exist