# Limits – Problems

### 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