Let f(x) be a function defined on some open interval that contains the number a except possibly at itself. then we say that the limit of f(x) on x approaches a is L, and we write

\(\lim_{x \to a} f(x) = L\)

If for every number ε > 0 there is a number δ > 0 such that if 0 < |x – a| < δ then |f(x) – L| < ε.

**Example 3: **Use the graph of f(x) = \(\sqrt{x}\) to find a number δ such that if |x – 4| < δ then \(|\sqrt{x}-2| < 0.4\).

Solution:

Here limit of the function f(x) = \(\sqrt{x}\) at x = 4 is 2 i.e. L = 2 and a = 4, ε = 0.4

Now

-0.4 < \(\sqrt{x}\) – 2 < 0.4

1.6 < \(\sqrt{x}\) < 2.4 whenever |x – 4| < δ

We need to concentrate near region of the point (4,2)

We need to find the values of x for which the curve y = \(\sqrt{x}\) lies between the horizontal line y= 1.6 and y = 2.4. Therefore, we graph the curve y = \(\sqrt{x}\), y = 1.6 and y = 2.4 near point (4,2).

Now the point of intersection of y = \(\sqrt{x}\) and y =1.6 is x = 2.56

Again the point of intersection of y = \(\sqrt{x}\) and y = 2.4 is x = 5.76

Now rounding to be safe we have if 2.56 < x < 5.76 then 1.6 < \(\sqrt{x}\) < 2.4

This interval (2.56, 5.76) is not symmetric about x = 4. The distance from x = 4 to the left end point is 4 – 2.56 = 1.44 and the distance of x = 4 from the right end point.

x = 5.76 is 5.76 – 4 = 1.76

Therefore, δ = smallest of two numbers = 1.44

Hence we can write |x – 4| < 1.44 then |\(\sqrt{x}\) – 2| < 0.4

## Continuity

**Definition: **A function f(x) is continuous at x = a if \(\lim_{x \to a} f(x) = f(a)\) [Functional Value = Limiting Value]

Precisely, a function f(x) is continuous at x = a if

- f(a) is defined (i.e. a is in the domain of
*f*) - \(\lim_{x \to a} f(x)\) exists
- \(\lim_{x \to a} f(x) = f(a)\)

**Example: **Use Continuity to evaluate limit.

\(\lim_{x \to 4} \frac{5 + \sqrt{x}}{\sqrt{5 + x}}\)

Solution:

Since the function \(f(x) = \frac{g(x)}{h(x)}\) is being a quotient of two continuous functions g(x) = 5 + \(\sqrt{x}\) and h(x) = \(\sqrt{5 + x}\) everywhere in their domain. In particular at x = 4 and hence the quotient function f(x) is also continuous at x = 4

\(\lim_{x \to 4}f(x) = f(4)\) \(\frac{5 + \sqrt{4}}{\sqrt{5 + 4}} = \frac{7}{3}\)

### Horizontal Asymptotes

The line y = L is called a horizonal asymptotes of the curve y = f(x) if either

\(\lim_{x \to ∞} f(x) = L \enspace or \enspace \lim_{x \to -∞} f(x) = L\)

**Example: **Find the horizontal asymptotes of curve y = tan^{-1}x

Solution:

Now

\(\lim_{x \to ∞} tan^{-1}x = tan^{-1}(∞) = \frac{\pi}{2}\)Therefore, y = \(\frac{\pi}{2}\) is a horizontal asymptotes

Again

\(\lim_{x \to -∞} tan^{-1}x = tan^{-1}(-∞) = -\frac{\pi}{2}\)which shows that y = –\(\frac{\pi}{2}\) is another horizontal asymptotes of same function y = tan^{-1}x.

### Vertical Asymptotes

Let f(x) be a given function. If there exists a number a such that any one of the following is true

- \(\lim_{x \to a} f(x) = ∞ \enspace or \enspace -∞ \)
- \(\lim_{x \to a^{-}} f(x) = ∞ \enspace or \enspace -∞ \)
- \(\lim_{x \to a^{+}} f(x) = ∞ \enspace or \enspace -∞ \)

Then x = 0 is called the vertical asymptotes.

**Example:** Find the vertical Asymptotes of \(y = \frac{x + 2}{x^2 + 2x – 8}\).

Solution:

We will get the y = ∞ when the denominator part is zero. So,

*x*^{2} + 2*x* – 8 = 0

(*x* + 4)(*x* – 2) = 0

*x* = –4 or *x* = 2

So, y = -4 and x = 2 are the vertical asymptotes.