Limit Formula

Limits help us comprehend how functions behave as their inputs approach certain values. Think of a limit as the destination that a function aims to reach as the input gets closer and closer to a specific point.

In this article, we will explore the essential limit formulas that form the backbone of calculus. These formulas are like the rules of a game, guiding us on how to find limits in various scenarios. Whether you’re adding, subtracting, multiplying, or dividing functions, there are specific formulas to help you determine the limit.

A limit is the value that a function (or sequence) “approaches” as the input (or index) approaches some value.

Mathematically, for a function f(x), the limit of f(x) as approaches c is L, written as:

lim⁡_x→cf(x) = L

This means that for every ϵ > 0, there exists a δ>0 such that whenever 0 < ∣x − c∣ < δ, it follows that ∣f(x) − L∣ < ϵ.

Read More about Formal Definition of Limits.

There are various formulas involving the limit such as:

• Basic Limit Formulas
• Trigonometric Limits
• L-hospital Rule
• Exponential Limits
• Logarithmic limits

Let’s discuss these formulas in detail as follows:

Basic Limit Formulas

Trigonometric Limits

To evaluate trigonometric limits, we have to reduce the terms of the function into simpler terms or into terms of sinθ and cosθ.  

• lim_{x ⇢ 0}  \frac{sinx}{x}  = lim_{x ⇢ 0} \frac{x}{sinx}  = 1
• lim_{x ⇢ 0} tanx/x = lim_{x ⇢ 0} x/tanx =1

As we considered our first one, 

lim_{x ⇢ 0} sinx/x =1      

Using L-Hospital

lim_{x ⇢ 0} cosx/1 

lim_{x ⇢ 0} cos(0)/1 = 1/1 =1

If the function gives an indeterminate form by putting limits, Then use the l-hospital rule.

Indeterminate  Form 

0/0, ∞/∞, ∞-∞, ∞/0, 0^∞, ∞⁰ , 0⁰, ∞^∞

L-hospital Rule

If we get the indeterminate form, then we differentiate the numerator and denominator separately until we get a finite value. Remember we would differentiate the numerator and denominator the same number of times. Similarly for all trigonometric function,

• lim_{x ⇢ 0} sin^-1x/x = lim_{x ⇢ 0} x/sin^-1x = 1

lim_{x ⇢ 0} sin^-1x/x =1

lim_{x ⇢ 0} 1/√1+x²  [Using L-Hospital] 

= 1/√(1 + (0)²) = 1

• lim_x⇢0 \frac{tan^{-1}x}{x}  =1
• lim_{x ⇢ a}  sin x^o/x = π/180
• lim_x⇢0 cosx = 1
• lim_{x ⇢ a}  sin(x-a) / (x-a) =1

lim_{x ⇢ a} sin(x – a) / (x – a) 

=1

lim_{x ⇢ a} cos(x – a)/1 

= lim_{x ⇢ a} cos(a – a) = cos(0) =1

• lim_x⇢∞ sinx/x = 0
• lim_x⇢∞ cosx/x = 0
• lim_x⇢∞ sin(1/x) / (1/x) =0

lim_{x ⇢ ∞} sin(1/x)/(1/x) = 0

Let 1/x = h

So, limits changes to 0

Because 1\∞ = 0

lim_{h ⇢ 0} sinh/h

As we see before, If lim_{x ⇢ 0} sinx/x = 1

So, lim_{h ⇢ 0} sinh/h = 1

Read More about L’Hospital Rule.

Exponential Limits

Some of the common formula related to exponents are:

• lim_{x ⇢ 0}  e^x – 1 /x = 1
• lim_{x ⇢ 0}  a^x – 1 /x = log_ea
• lim_{x ⇢ 0} e^λx – 1 /x = λ

Read More about Exponents.

Logarithmic Limits

Some of the common formula related to logarithmic limits are:

• lim_{x ⇢ 0} log(1 + x) /x = 1
• lim_{x ⇢ e} log_ex = 1
• lim_{x ⇢ 0} log_e(1 – x) /x  = -1
• lim_{x ⇢ 0} log_a(1 + x) /x = log_ae

Some of the important results involving limits are:

1. lim_x⇢0 (1+x)^{\frac{1}{x}}  = e
2. lim_x⇢0 \left( 1+\frac{1}{x} \right)^x  = e
3. lim_x⇢0 \frac{e^x-1}{x} =1
4. lim_x⇢0\frac{a^x-1}{x}  = log_ea
5. lim_x⇢0 \frac{1-cosmx}{x^2}  = m²/2
6. lim_x⇢0 \frac{1-cosmx}{1-cosnx} = \frac{m^2}{n^2}

Some Shortcut Formulas

1. lim_x⇢0 \left( 1+ \frac{a}{b}x \right)^\frac{c}{dx} = e^\frac{ac}{bd}
2. lim_x⇢0 \left( 1+ \frac{a}{bx} \right)^\frac{cx}{d} = e^\frac{ac}{bd}
3. lim_x⇢a f(x)^g(x) = e^{limx\to a\left[ f(x)-1 \right].g(x)}

Question 1: Solve, lim_x⇢0  (x – sinx ) /(1 – cosx).

Solution:

Using L-hospital,

lim_{x ⇢ 0} (1 – cosx) / (sinx)

lim_{x ⇢ 0} sinx / cosx = sin(0) / cos(0) = 0/1 = 0

Question 2: Solve, lim_{x ⇢ 0} (e^2x -1) / sin4x.

Solution:

Using L-hospital 

lim_{x ⇢ 0} (2)(e^2x) / cos4x

lim_{x ⇢ 0}  2(e⁰) / cos4(0) = 2/1= 2

Question 3: Solve, lim_{x ⇢ 0} (1 – cosx) / x²

Solution:

Using L-hospital 

lim_{x ⇢ 0} sinx /2x = 1/2 {sinx/x = 1}

Question 4: Solve, lim_{x ⇢ ∞} \frac{x+sinx}{x}

Solution:

lim_{x ⇢ ∞} (1 + \frac{sinx}{x}        )

1 + lim_{x ⇢ ∞} \frac{sinx}{x}

As we know, x = ∞  

So 1/x = 0

1 + lim\frac{1}{x}⇢∞ \frac{sin\frac{1}{x}}{x}

1 + 0 = 0                                                                                                                                                 

Question 5: Solve, lim_{x ⇢ π/2} (tanx)^cosx 

Solution:

let Y = lim_{x ⇢ π/2}  (tanx)^cosx

Taking log_e both sides,

 log_eY = lim_{x ⇢ π/2}  log_e(tanx)^cosx 

 log_eY = lim_{x ⇢ π/2}  cosx log_e(tanx)

log_ey = lim_{x ⇢ π/2} loge(tanx)/secx

Using l-hospital,

log_ey = lim_{x ⇢ π/2} cosx /sin²x = 0

Now, taking exponent on both sides,

Y = lim_{x ⇢ π/2} e⁰ 

Y = lim_{x ⇢ π/2}  (tanx)^cosx = 1  

Question 6: lim_{x ⇢ 0}  \frac {e^x -(1+x+ \frac {x^2}{2})}{ x^3}

Solution:

limx⇢0 \frac{1+\frac{x}{1!} + \frac{x2}{2!} + \frac{x3}{3!} – ( 1+ x+ \frac{x2}{2!} ) }{x3}

limx⇢0 \frac{\frac{x3}{3!}}{x3}  = 1/3! =1/6

Question 7: Solve, lim_{a ⇢ 0}  \frac{x^a-1}{a}

Solution:

Using l-hospital (Differentiating numerator and denominator w.r.t a)  

lim_{a ⇢ 0}  x^alogx = logx

Question 8: Solve, lim_{x ⇢ 0} \frac{x^2+x-sinx}{x^2}

Solution:

lim_{x ⇢ 0} \frac{x^2+x-(x-x^3/3!)}{x^2}

lim_{x ⇢ 0} 1 + x/3! = 1

Problem 1: Simplify.

• lim_x→2(3x + 4)
• lim_x→-1(x² + 2x + 1)
• lim_x→0(sin²x/2x)
• lim_x→3[(x² – 9)/(x – 3)]
• lim_x→∞[(5x + 7)/(2x – 3)]
• lim_x→0[(1 – cos x)/x]
• lim_x→-∞(2x³ − x² + x)
• lim_x→∞(e^-x)
• lim_x→0[(e^x – 1)/x]
• lim_x→∞[(2x² − 5x + 3​)/(x – 3)]

What is a limit in calculus?

A limit in calculus describes the value that a function approaches as the input (or variable) approaches a certain point.

How do you evaluate a limit?

Limits can be evaluated using various methods, such as direct substitution, factoring, rationalizing, using special limit formulas, and applying the squeeze theorem or L’Hôpital’s rule.

What are some basic limit formulas?

Some basic limit formulas include:

What is the limit of a constant?

The limit of a constant k as x approaches any value is the constant itself i.e., lim_x→c​k = k.

What is the Squeeze Theorem?

The Squeeze Theorem states that if f(x) ≤ g(x) ≤ h(x) for all x near c (except possibly at c) and lim_x→c​f(x) = lim_x→ch(x) = L then

lim_x→cg(x) = L