5.10 Limits (Page 5/5)

Functions

Page 5 / 5

$0 / 0, \infty / \infty, 0. \infty, \infty - \infty, 0^{0,} \infty^{0,} 1^{\infty}$

Again, we emphasize to distinguish interpretation with each of the indeterminate limit forms given above. In order to clarify the point further, let us consider two limits :

$\lim_{x \to π / 2}^{} \sin x^{\tan x}$ $\lim_{x \to π / 2}^{} 1^{\tan x} = 1$

In the first case, the base is approaching 1 and exponent is approaching infinity. Hence, it is indeterminate limit form. In the second case, base is 1 – not approaching to 1. Hence, it is not an indeterminate form and is evaluated to 1.

Evaluation of limit

There are three distinct regimes based on the discussion as above. We evaluate limit in accordance with following algorithm :

1: The function is not in indeterminate function form.

We simply plug in the value of test point into the function. The function value is equal to the limit of function.

2: The function is in indeterminate function form.

We transform the functions into determinate form. There are many techniques to transform an indeterminate form. Rationalization, simplification etc are important means to change forms. Expansion series of transcendental functions are also helpful. Besides, there are forms whose limits are known. We attempt to structure given expression in those standard forms and then find the limit. Finally, there are specific algorithms depending on nature of function, which help to remove indeterminate form and find limit. We shall study specific techniques in specific context.

3: The function is piecewise defined.

Using two approaches outlined above, we determine left and right limit and see whether they are equal or not? If equal, then limit is equal to either of left and right limits.

Note : We shall divide evaluation of limit in separate categories for different function types. The evaluations of limits shall be dealt separately in detail. Here, we work with few fundamental limits only.

Example

Problem : Plot the graph of function and determine limit when x->1.

$f (x) = \frac{x + 3}{x - 1}$

Solution : Here, singularity exists at x=1. The function is not defined at this point. Rearranging, we have :

$\frac{x + 3}{x - 1} = \frac{x - 1 + 4}{x - 1} = 1 + \frac{4}{x - 1}$

Here core graph is 4/x graph. We obtain graph of 4/(x-1) by shifting graph of 4/x towards right by 1 unit. In order to obtain the graph of given function, we shift the graph of 4/(x-1) up by 1 unit. We see that the function has one zero at x=-3. It has one asymptote at x=1. On the other hand, the y-intercept of function is -3.

Graph of rational function

Left hand limit is :

$\Rightarrow \lim_{x \to 1 -}^{} 1 + \frac{4}{x - 1} = - \infty$

Right hand limit is :

$\Rightarrow \lim_{x \to 1 +}^{} 1 + \frac{4}{x - 1} = \infty$

Since left and right hand limits are not equal, the limit of function does not exist as x approaches to 1.

Exercises

Determine limit :

$\Rightarrow \lim_{n \to \infty}^{} \frac{n}{n + 1}$

The limit has ∞/∞ indeterminate form. Dividing each term of numerator and denominator by n,

$\Rightarrow \frac{n}{n + 1} = \frac{\frac{n}{n}}{\frac{n}{n} + \frac{1}{n}} = \frac{1}{1 + \frac{1}{n}}$

As n→∞, 1/n→0. Hence,

$\Rightarrow \lim_{n \to \infty}^{} \frac{n}{n + 1} = 1$

Determine limit :

$\lim_{x \to 0}^{} \cos x$

The limit has determinate form. Since cosx is a continuous function, its limit is equal to its value at x=0 i.e. cos0 = 1. Hence,

$\Rightarrow \lim_{x \to 0}^{} \cos x = 1$

Determine limit :

$\lim_{x \to 0}^{} \log_{0.5} x$

The limit has determinate form. Here, $\log_{0.5} x$ is a continuous function. The base of exponential function is less than 1. As x approaches zero, the function approaches positive infinity (refer its graph).

$\Rightarrow \lim_{x \to 0}^{} \log_{0.5} x = \infty$

Determine limit :

$\lim_{x \to 0}^{} \frac{1}{x - 1}$

Note that we are not testing limit at singularity. The function is determinate form. Hence, limit is :

$\Rightarrow \lim_{x \to 0}^{} \frac{1}{x - 1} = \frac{1}{0 - 1} = - 1$

Determine limit :

$\lim_{x \to \infty}^{} \frac{x^{2}}{x^{2}}$

The limit has ∞/∞ indeterminate form. Simplifying,

$\Rightarrow \lim_{x \to \infty}^{} 1 = 1$

Determine limit :

$\lim_{x \to 0}^{} \frac{| x^{3} |}{x}$

The limit has 0/0 indeterminate form. For $x < 0, | x^{3} | = - x^{3}$ . Hence,

$\Rightarrow \frac{| x^{3} |}{x} = - \frac{x^{3}}{x} = - x^{2}$

$\Rightarrow \lim_{x \to 0 -}^{} - x^{2} = 0$

For $x > 0, | x^{3} | = x^{3}$ . Hence,

$\Rightarrow \frac{| x^{3} |}{x} = \frac{x^{3}}{x} = x^{2}$ $\Rightarrow \lim_{x \to 0 +}^{} x^{2} = 0$

Clearly, $L_{l} = L_{r} = L$ . Thus,

$\Rightarrow \lim_{x \to 0}^{} \frac{| x^{3} |}{x} = 0$

Determine limit :

$\lim_{x \to 1}^{} \frac{(x^{2} - 1)}{| x - 1 |}$

The limit has 0/0 indeterminate form. For $x < 1, | x - 1 | = - (x - 1)$ . Hence,

$\frac{(x^{2} - 1)}{| x - 1 |} = \frac{(x^{2} - 1)}{- (x - 1)} = - (x - 1)$ $\Rightarrow \lim_{x \to 1}^{} \frac{(x^{2} - 1)}{| x - 1 |} = \lim_{x \to 1}^{} - (x + 1) = - 2$

For $x > 1, | x - 1 | = (x - 1)$ . Hence,

$\frac{(x^{2} - 1)}{| x - 1 |} = \frac{(x^{2} - 1)}{(x - 1)} = (x - 1)$ $\Rightarrow \lim_{x \to 1 +}^{} \frac{(x^{2} - 1)}{| x - 1 |} = \lim_{x \to 1 +}^{} (x + 1) = 2$

Clearly, $L_{l} \neq L_{r}$ . Hence, given limit does not exists.

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Source: OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64

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