Evaluating Limit

Limit is the basic theory of Calculus and Analysis. The limit of a function at a point xin its domain is a value that a function approaches as its argument approaches $x$. In other words, a function is said to have a Limit L at $x$ if it is possible to make the function arbitrarily close to $L$ by selecting the value closer and closer to $x$. Note that the actual values are irrelevant to the value of limit. Mathematically, the limit of a function is represented as:

$\lim_{x \to k} f(x) = L$

The limit of a function is read as “ Limit of $f(x)$ as $x$ approaches $k$ is $L$”.

Evaluating Limits means to determine the value that the function is approaching at a certain point. When evaluating limits, we first check to see if the function is continuous. If we find that the limit is continuous at the point where we are evaluating it, we simply substitute the value and solve the function.

In this article, we will discuss how to find the limit of a function using different evaluation methods.

(Image will be uploaded soon)

How to Evaluate Limit?

Let us learn different methods on how to evaluate limits:

Limit by Factoring

Limit by factoring is a method for evaluating limits that require determining and evaluating common factors.

Let us understand how to calculate limit by factoring with an example:

Evaluate $\lim_{y \to 5} \dfrac{y^2 - y - 20}{y - 5}$

Solution:

We have,

$\lim_{y \to 5} \dfrac{y^2 - y - 20}{y - 5} = \lim_{y \to 5} \dfrac{(y - 5)(y + 4)}{y - 5}$

$\Rightarrow \lim_{y \to 5}( y + 4)$

$\Rightarrow 9$

Limit by Substitution

A limit is defined as the value of a function approaches as the variable within that function gets closer and closer to a specified value. Suppose, we have a limit $\lim_{x \to k}f(x)$. This represents the value of $f(x)$ when $x$ is closer to $k$ but not exactly equals to $k$. The substitution rule determines the limit by simply substituting $x$ with $k$. Mathematically, this rule is defined as:

$\lim_{y \to k}f(x) = f(k)$

Let us understand with an example:

Evaluate the following limit using substitution rule:

$\lim_{y \to -1}(y^2 + y - 6)$

Solution:

The limit can be evaluated easily. Simply, use the substitution rule and substitute $y = -1$ and we have:

$\lim_{y \to -1}(y^2 + y - 6) = (-1)^2 + (-1) - 6 = - 6$

Remember: The substitution rule is not always valid. To use the substitution rule, the function $f(x)$ must be continuous.

L’ Hospital Rule

L’s Hospital Rule is the method of evaluating limit of certain quotient by means of derivatives. Specifically, under certain cases, it enables us to replace lim $\dfrac{f(x)}{g(x)}$ by $\dfrac{f^{\prime}(x)}{g^{\prime}(x)}$, which is quite easier to evaluate.

L’s hospital rule states that,

If $f$ and $k$ are differentiable functions such that:

• $k^{\prime}(x)= 0$ on an open interval $I$ contain $y$.

• $\lim_{x \to y}f(x)= 0$ and $\lim_{x \to y}k(x)= 0$ or $\lim_{x \to y}f(x)= \pm \infty$ and $\lim{x \to y}k(x)= \pm \infty$

• $\lim_{x \to y} \dfrac{f^{\prime}(x)}{k^{\prime}(x)}$ exists

Then,

$\lim_{x \to y} \dfrac{f(x)}{k(x)} = \lim_{x \to y}\dfrac{f^{\prime}(x)}{k^{\prime}(x)}$

Evaluate the following limit using L’s hospital rule:

$\lim_{y \to \infty}\dfrac{In(y)}{y^{\frac{1}{3}}}$

Solution:

As $In(y) \to \infty$ and $y^{\frac{1}{3}} \to \infty$ as $y \to \infty$, we can use the L’s hospital rule

$\lim{y \to \infty} \dfrac{In(y)}{y^{\frac{1}{3}}} = \dfrac{\dfrac{1}{y}}{\dfrac{1}{3}y^{-\frac{2}{3}}}$

$\Rightarrow \lim{y \to \infty} \dfrac{3}{y^\frac{1}{3}} = 0$

Limit at Infinity 

If $k$ is a positive real number, then

$\lim_{x \to \infty} \dfrac{1}{x^n} = 0$ Limit Towards Right

Further, if $x^n$ is defined when $x < 0$, then

$\lim_{x \to \infty} \dfrac{1}{x^n} = 0$ Limit Towards Left

Let us understand limit at infinity with an example:

Evaluate the following limit:

$\lim_{y \to \infty}\dfrac{2y^2 - 3y + 7}{y^2 + 47y + 1}$

Solution:

As y approaches to the infinity both numerator and denominator approaches to y².

$\lim_{y \to \infty} \dfrac{2y^2 - 3y + 7}{y^2 + 47y + 1} = \lim_{y \to \infty} \dfrac{2y^2 - \dfrac{3}{y} + \dfrac{7}{y^2}}{y^2 + \dfrac{4y}{y} + 1}$

As $y$ approaches infinity, all the quotients with similar power of $z$ in the denominator approaches 0, leaving 32 in numerator and 1 in the denominator. Therefore, again the limit is 2.

$\lim_{y \to infty}\dfrac{2y^2 - 3y + 7}{y^2 + 47y + 1} = 2$

Evaluating Limit By Rationalising

Let us learn how to find limit calculus by rationalising. We can find the limit of some function by some rationalising techniques. In rationalising technique, we rationalise the numerator of a function. Rationalising the numerator means multiplying the numerator and denominator by the conjugate of the numerator. For example, the conjugate of $\sqrt{x} +7$ is $\sqrt{x} -7$.

Evaluate the following limit by rationalising:

$\lim_{y \to 0}\dfrac{ \sec y - 1}{y^2}$

Solution:

$\lim_{y \to 0}\dfrac{\sec y - 1}{y^{2}} = \lim_{y \to 0}\dfrac{(sec y - 1)( sec y + 1)}{y^{2} ( sec y + 1)}$

$\lim_{y \to 0}\dfrac{\sec y - 1}{y^2} = \lim_{y \to 0}\dfrac{(sec y - 1)( sec y + 1)}{y^2 ( sec y + 1)}$

$\Rightarrow \lim_{y \to 0}\dfrac{tan^2y}{y^2 ( sec y + 1)}$

$\Rightarrow \lim_{y \to 0}\left(\dfrac{\sin y}{y}\right)^2 \left(\dfrac{1}{\cos^2 y}\right) \left(\dfrac{1}{\sec y + 1}\right)$

$\Rightarrow 1^2 \times \dfrac{1}{1} \times \dfrac{1}{2} = \dfrac{1}{2}$

One - Sided Limit

One -sided limit considers only those values of a function that approaches a value either from above or below.

The right side of a limit of a function $f$ as it reaches $k$ is the limit $\lim_{x \to k^{+}} f(x) = L$

The left side of a limit of a function $f$, $\lim_{x \to k^{-}} f(x) = L$

The notation “$x \to k^{-}$” and “$x \to k^{+}$” considers only those values of $x$, tha are less than or greater than $k$ respectively.

Conjugates

If you try substitution and get $\dfrac{0}{0}$ ( 0 divided by 0) and the expression contains a square root in it, then rationalize the expression as you rationalize in Algebra. That is, multiply the numerator and denominator by the conjugate of the part that contains a square root in it.

Let us learn how to find limit using conjugates technique with an example:

Evaluate the following limit using Conjugate rule:

$\lim_{y \to 0} \dfrac{\sqrt{1+y} - 1}{y}$ 

Solution:

As the direct substitution gives the indeterminate form $\dfrac{0}{0}$, we will multiply both the numerator and denominator by the conjugate of numerator $\sqrt{1+y}+1$:

$\lim_{y \to 0} \dfrac{\sqrt{1+y}-1}{y} = \lim_{y \to 0} \dfrac{(\sqrt{1+y}-1)-\sqrt{1+y}+1}{y(\sqrt{1+y}+1)}$

$\Rightarrow \lim_{y \to 0} \dfrac{\sqrt{1+y}-1}{y(\sqrt{1+y}+1)}$

$\Rightarrow \lim_{y \to 0} \dfrac{y}{y(\sqrt{1+y}+1)}$

$k = \dfrac{1}{2}$

Laws of Limit

Following are the laws of limit:

Assumption: $c$ is constant and $\lim_{x \to a}f(x)$ and $\lim_{x \to a}k(x)$ exists

1. $\lim_{x \to a}k = k$ - The limit of subtraction is equal to the subtraction of the limits.

2. $\lim_{x \to k}x = k$ - The limit of linear function is equialaitbent to the number $x$ is approaching.

3. $\lim_{x \to k}x^n = k^n$, where $n$ is a positive integer.

4. $\lim_{x \to k} \sqrt[n]{x} = \sqrt[n]{k}$, where $n$ is a positive integer, and if $n$ is even we assume that $k > 0$.

5. $\lim_{x \to k} \sqrt[n]{f(x)} = \sqrt[n]{\lim{x \to k} f(x)}$, where $n$ is a positive integer , and if $n$ is even we assume that $\lim_{x \to k}f(x) > 0$.

6. $\lim_{x \to k}ck(x) = c \lim_{x \to k} f(x)$.

7. $\lim_{x \to a}[f(x) + k(x)] = \lim_{x \to a}f(x) + \lim_{x \to a}k(x)$- The limit of addition is equal to the addition of the limits.

8. $\lim_{x \to a}[f(x) - k(x)] = \lim_{x \to a}f(x) - \lim_{x \to a}k(x)$- The limit of subtraction is equal to the subtraction of the limits.

9. $\lim_{x \to a}[f(x) + k(x)] = \lim_{x \to a}f(x) \cdot \lim_{x \to a}k(x)$- The limit of the product is equal to the subtraction of the limits.

10. $\lim_{x \to a}\dfrac{f(x)}{k(x)} = \dfrac{\lim_{x \to a}f(x)}{\lim_{x \to a}k(x)}$ ( If $\lim_{x \to k} k \neq 0$- The limit of the quotient is equal to the quotient of the limits.

Solved Example:

1. Evaluate the following limit:

$\lim_{y \to \infty} \dfrac{y^2+2y+4}{3y^2+4y+125345}$

Solution:

$\lim_{y \to \infty} \dfrac{y^2+2y+4}{3y^2+4y+125345} = \lim_{y \to \infty} \dfrac{y^2+\dfrac{2}{y}+\dfrac{4}{y^2}}{3+\dfrac{4}{y}+\dfrac{125345}{y^2}} = \dfrac{1+0+0}{3+0+0} = \dfrac{1}{3}$

2. $\lim_{y \to 1}\dfrac{y^2 - 1}{y -1}$

Solution:

By factoring $(y^2 - 1)$ into $(y-1)(y+1)$, we get:

$\lim_{y \to 1}\dfrac{y^2 - 1}{y - 1} = \lim_{y \to 1}\dfrac{(y - 1)(y + 1)}{y - 1}$

$\Rightarrow \lim_{y \to 1}(y -1)$

Now, we can substitute $y =1$ to the limit

$\lim_{y \to 1}(y-1) = (1+1) = 2$