How to solve the problems of limits?

In calculus, limits are very essential for solving the various kinds of problems at a particular point. Differential calculus & integral calculus can be defined by using the limits. The value of a function can be finite or infinite by putting the specific point.

By using the types & rules of limits, all the complex problems of this topic can be solved easily. The sum, difference, product, & quotient rules of limits find the finite result of the function. if the function gives an indeterminate or undefined result then L’hopital’s rule is used.

In this post, we will discuss all the basics of limits along with the solved examples.

What are the limits in calculus?

A limit is a kind of calculus that is used to find the numerical result of the function at a particular point. It is defined as a value that approaches as the given input of the function becomes nearer to a particular number.

The equation of the limit is written as:

$\lim_{t \to u} h(t) = S$

Where "lim" is the notation of limit, h(t) is the given function, t is the variable of limits, u is the specific point, and S is the numerical result of the function at u.

The numerical result of limits can be determined by placing the corresponding term of the function (specific point). The specific point of limits is only applied to the corresponding variables.

Rules of limits

Here are some well-known rules of limits

Name of the ruleRule
Power rule $\lim_{t \to u}h^n(t)=[\lim_{t \to u}h(t)]^n$
Sum rule$\lim_{t \to u}[h(t)+g(t)]=\lim_{t \to u}h(t)+\lim_{t \to u}g(t)$
Product rule$\lim_{t \to u}[h(t)\times g(t)]=\lim_{t \to u}h(t)\times \lim_{t \to u}g(t)$
Constant rule $\lim_{t \to u}L = L$, where L is any constant
Constant multiplier rule$\lim_{t \to u}L\times h(t) = L\times\lim_{t \to u}h(t)$, where L is any constant
Difference rule$\lim_{t \to u}[h(t)- g(t)]=\lim_{t \to u}h(t)- \lim_{t \to u}g(t)$
Quotient rule$\lim_{t \to u}\frac{h(t)}{g(t)}=\frac{\lim_{t \to u}h(t)}{\lim_{t \to u}g(t)}$
L’hopital’s rule$\lim_{t \to u}\frac{h(t)}{g(t)}=\frac{\lim_{t \to u}h'(t)}{\lim_{t \to u}g'(t)}$

Types of limits in calculus

There are 3 basic types of limits in calculus.

   The left-hand limit (L-H-L) in calculus

   The right-hand limit (R-H-L) in calculus

   The two-sided limit (T-S-L) in calculus

1. Left-hand limit

In calculus, the L-H-L of a function is the value or number of the function that approaches when the variable approaches its limit form the left-hand side of the interval. The L-H-L is denoted by the negative sign above the specific point. The equation of L-H-L is:

$\lim_{t \to u^{-}} h(t) = S$

Let us take an example to understand L-H-L.

Example

Evaluate $h(t) = 6t^4 + 5t^3 - 4^t + 16$ as "t" approaches to 4.

Solution

Step I: Identify the given function, variable, & specific point of limits & apply the limit notation. 

$h(t) = 6t^4 + 5t^3 - 4t + 16$

variable of function = t

specific point = u = 4

$\lim_{x \to 4^{-}}h(t) = \lim_{x \to 4^{-}}(6t^4 + 5t^3 - 4t + 16)$

Step II: Apply the limit notation separately to each function by using the sum & difference rules of limits.

$\lim_{x \to 4^{-}}(6t^4 + 5t^3 - 4t + 16)=\lim_{x \to 4^{-}}(6t^4) + \lim_{x \to 4^{-}}(5t^3) - \lim_{x \to 4^{-}}(4t) + \lim_{x \to 4^{-}}(16)$

Step III: Take the constant coefficients outside the limit notation by using the constant multiplier rule.

$\lim_{x \to 4^{-}}(6t^4 + 5t^3 - 4t + 16)=6\lim_{x \to 4^{-}}(t^4) + 5\lim_{x \to 4^{-}}(t^3) - 4\lim_{x \to 4^{-}}(t) + \lim_{x \to 4^{-}}(16)$

Step IV: Substitute the specific point "4" in the place of variable "t" in the above expression by using the power & constant rules to get the numerical result of the given function.

$\lim_{x \to 4^{-}}(6t^4 + 5t^3 - 4t + 16)=6(4^4) + 5(4^3) - 4(4) + (16)=6(256) + 5(64) – 4(4) + (16)=1536 + 320 – 16 + 16=1826$

2. Right-hand limit (R-H-L)

In calculus, the R-H-L is value of a function in which the limit of that function as it approaches form the right-hand side of the interval. The R-H-L is denoted by the negative sign above the specific point. The equation of R-H-L is:

$\lim_{t \to u^{+}} h(t) = S$

Let's take an example of R-H-L to understand this type of limit accurately.

Example

Evaluate $h(t) = 12t^5 + 2\sin(t) + 15t - 4$ as "t" approaches to 2.

Solution

Step I: Identify the given function and apply the limit notation.

$\lim_{t \to u^{+}} h(t) = \lim_{t \to 2^{+}}(12t^5 + 2\sin(t) + 15t - 4)$

Step II: Apply the limit notation separately to each function by using the sum & difference rules of limits and take the constant coefficient outside the limit notation.

$\lim_{t \to 2^{+}}(12t^5 + 2\sin(t) + 15t - 4)=\lim_{t \to 2^{+}}(12t^5) + \lim_{t \to 2^{+}}(2\sin(t)) + \lim_{t \to 2^{+}}(15t) - \lim_{t \to 2^{+}}(4)$
$=12\lim_{t \to 2^{+}}(t^5) + 2\lim_{t \to 2^{+}}(\sin(t)) + 15\lim_{t \to 2^{+}}(t) - \lim_{t \to 2^{+}}(4)$

Step III: Substitute the specific point "2" in the place of variable "t" in the above expression.

$=12(2^5)+ 2(\sin(2))+ 15(2) - 4=384 + 2\sin(2) + 30 - 4=410 + 2\sin(2)$

The above problem of right-hand limit can also be solved by using a limit calculator to get avoid such a lengthy solution.

Note:

Right-hand limit is not always equal to left-hand limit.

For example: $\lim_{x\to0^{+}}\frac{|\sin(x)|}{\sin(x)} = 1$, but $\lim_{x\to0^{-}}\frac{|\sin(x)|}{\sin(x)} = -1$

3. Two-sided limit (T-S-L)

If the value of limits comes from both the direction (positive or negative) is said to be the right-hand limit. In simple words, if the left-hand & right-hand limits holds for a function, then that function is said to be the two-sided limit.

The equation of two-sided limit is same as the simple equation of limits.

$\lim_{t \to u} h(t)=S$

Below is a solved example of two-sided limit.

Example

Evaluate "$4t^2 – 8t^5 + t + 12x$", when "t" approaches 1.

Solution

Step I: Identify the given function & apply the limit notation.

$h(t)=4t^2 – 8t^5 + t + 12x$

$\lim_{t \to u}h(t)=\lim_{t\to1}(4t^2 – 8t^5 + t + 12x)$

Step II: Apply the limit notation separately to each function.

$\lim_{t\to1}(4t^2 – 8t^5 + t + 12x)=\lim_{t\to1}(4t^2) – \lim_{t\to1}(8t^5) + \lim_{t\to1}(t) + \lim_{t\to1}(12x)$

Step III: Take the constant coefficients outside the limit notation by using the constant multiplier rule.

$\lim_{t\to1}(4t^2 – 8t^5 + t + 12x)=4\lim_{t\to1}(t^2) – 8\lim_{t\to1}(t^5) + \lim_{t\to1}(t) + 12\lim_{t\to1}(x)$

Step IV: Substitute the specific point "1" in the place of variable "t" in the above expression by using the power & constant rules to get the numerical result of the given function.

$\lim_{t\to1}(4t^2 – 8t^5 + t + 12x)=4(1^2)-8(1^5)+1+12(x)=-3+12x$

How to solve the problems of limits using L’hopital’s rule?

L’hopital’s rule is a general rule of limits used to find the limits of undefined functions by taking the differential of the numerator & denominator.

The equation of the L’hopital’s rule of limits is:

$\lim_{t\to u}\frac{h(t)}{g(t)}=\lim_{t\to u}\frac{\frac{d}{dt}h(t)}{\frac{d}{dt}g(t)}$

Example

Evaluate $\frac{4t^2 – 8t}{4t – 8}$ as "t" approaches 2.

Solution

Step I: Write the given function along with the limit notation & apply the difference & quotient rule.

$\lim_{t\to2}\frac{4t^2 – 8t}{4t – 8}=\frac{\lim_{t\to2}(4t^2 – 8t)}{\lim_{t\to2}(4t – 8)}=\frac{\lim_{t\to2}(4t^2) – \lim_{t\to2}(8t)}{\lim_{t\to2}(4t) – \lim_{t\to2}(8)}$

Step II: Substitute the value of the limit.

$=\frac{4(2^2) – 8(2)}{4(2) – 8}=\frac{0}{0}$

Step III: Use L’hopital’s rule of limits as the function gives an undefined result after substituting the limit value.

$\lim_{t\to2}\frac{4t^2 – 8t}{4t – 8}=\lim_{t\to2}\frac{\frac{d}{dt}(4t^2 – 8t)}{\frac{d}{dt}(4t – 8)} = \lim_{t\to2}\frac{8t – 8}{4 - 0}=\lim_{t\to2}(2t – 2)=2(2)-2=2$

Summary 

In this post, we have covered all the basics of how to solve the problems of limits. Now after reading the above post, you can solve any problem of limit either by using the rules or types of limits easily.


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