Limit is one of the important topics of calculus. When a function cannot be defined at some point, in such cases we can look at what the function is approaching to as it gets closer to that point. This is known as limit. There are times when the value of the function is same as the function limit. And at times they vary too.

$lim_{x -> a} h (x) = H$

This implies that as h (x) approaches to H as x gets closer to ‘a’.

This implies that there exist e > 0 such that h (x) lies in the interval (H – e, H + e) that is |h (x) - H| < e with existence of d > 0 such that x lies in either (a – d, a) or (a, a + d) that is 0 < |x – a| < d.

Consider the function $\frac{(5 x – 5 x^2)}{(x – 1)}$. This function is not defined at x = 1. So at this particular point we can take limit of this function and see as to what it approaches to when x -> 1.

That is $lim_(x -> 1)$ $\frac{(5 x – 5 x^2)}{(x – 1) }$

= $lim_(x -> 1)$ $\frac{5 x (1 – x)}{(x – 1) }$

= $lim_{x -> 1}$ (-5 x )

= - 5

It is not necessary that we use limits for only difficult functions. They can be used in simple functions too. Also at times we use limits sideways like a+ or a- that is right side of the point or left side of the point. This happens mostly in case of discontinuous functions. But the concept of limits solved a great term $\frac{1}{x}$ when x approaches to zero. With the help of limits we can say that as x gets closer to zero, $\frac{1}{x}$ approaches to zero.

Few important properties on limits are as follow:

**1)** $Lim_(x -> b)$ [m h (x)] = m $lim_(x -> b)$ h (x)

only if lim_(x -> b) h (x) $\neq$ 0

$lim_(x -> 1)$ [$\frac{(x^3 – 1)}{(x – 1)}$]

= $lim_(x -> 1)$ [$\frac{(x – 1) (x^2 + x + 1)}{(x – 1)}$]

= $lim_(x -> 1)$ ($x^2$ + x + 1)

= 1 + 1 + 1

= 3

**Example 2:** Evaluate $lim_{y \rightarrow 9}$ $\frac{3-\sqrt{y}}{9-y}$

**Solution:** Put y= 9 in $\frac{3-\sqrt{y}}{9-y}$, we get

$\frac{3-3}{9-9}$ = $\frac{0}{0}$ form

Reduced expression to avoid 0/0 form

Divide and multiply fraction by 3 + $\sqrt{y}$

$\frac{3-\sqrt{y}}{9-y}$ $\times$ $\frac{3+\sqrt{y}}{3+\sqrt{y}}$

= $\frac{(3^2-(\sqrt{y})^2}{(9-y)(3+\sqrt{y})}$

= $\frac{9-y}{(9-y)(3+\sqrt{y})}$

= $\frac{1}{(3+\sqrt{y})}$

By substituting y = 9 in above, we have

= $\frac{1}{(3+\sqrt{9})}$

= $\frac{1}{6}$

=> $lim_{y \rightarrow 9}$ $\frac{3-\sqrt{y}}{9-y}$ = $\frac{1}{6}$

$\frac{3-3}{9-9}$ = $\frac{0}{0}$ form

Reduced expression to avoid 0/0 form

Divide and multiply fraction by 3 + $\sqrt{y}$

$\frac{3-\sqrt{y}}{9-y}$ $\times$ $\frac{3+\sqrt{y}}{3+\sqrt{y}}$

= $\frac{(3^2-(\sqrt{y})^2}{(9-y)(3+\sqrt{y})}$

= $\frac{9-y}{(9-y)(3+\sqrt{y})}$

= $\frac{1}{(3+\sqrt{y})}$

By substituting y = 9 in above, we have

= $\frac{1}{(3+\sqrt{9})}$

= $\frac{1}{6}$

=> $lim_{y \rightarrow 9}$ $\frac{3-\sqrt{y}}{9-y}$ = $\frac{1}{6}$