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A limit is the value that a function or sequence approaches when the input approaches a certain value. Limits are used to define all important concepts in calculus and mathematical analysis like continuity, derivatives, and integrals. Let's look at the definition and representation of function limits, as well as certain rules, properties and examples of limits.
Also Read: Limits and Continuity
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What are Limits?
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Mathematical limits are unique real number. Consider the limit of a real-valued function "f" and a real number "c," which is generally defined as:
limx→c f(x)=L
It says, “The limit of f of x as x approaches c equals L.”
The "lim" denotes the limit, and the right arrow denotes the fact that function f(x) approaches the limit L as x approaches c.
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Continuity and Differentiability Detailed Video Explanation:
Limits and Function
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Consider the function f(x) = xsin x. As the value of x increases, this function does not approach any particular real number since we can always select a value of x to make f(x) greater than whatever number we want.
A function may reach one of two limits. One in which the variable approaches its limit by taking values that are greater than the limit, and the other in which the variable approaches its limit by taking values that are smaller.
The limit is not stated in this situation, although the right and left-hand limits do exist.
- When the limx→af(x)=A+ given the values of f near x to the right of a. The right hand limit of f(x) at an is stated to be this value.
- When the limx→af(x)=A− given the values of f near x to the left of a. The left hand limit of f(x) at an is referred to as this value.
- If and only if the left-hand limit equals the right-hand limit, the function's limit exists. limx→a−1f(x)=limx→a+f(x)= L
The limit of this function can be found between any two consecutive integers.
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Properties of Limit
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The following is a list of limit properties.
Let us assume that limx→af(x) and limx→ag(x) exist and c is a constant. Then,
- limx→a[c.f(x)] = c. limx→af(x)
- limx→a[f(x)±g(x)] = limx→af(x) ± limx→ag(x)
- limx→a[f(x)⋅g(x)] = limx→af(x)⋅limx→ag(x)
- limx→a[f(x).g(x)] = limx→af(x) . limx→ag(x)
- limx→a[f(x)/g(x)] = [lim x→af(x)] / [limx→ag(x)] provided limx→ag(x)≠0
As you can see, we only need to worry about the quotient limit if the denominator's limit is zero. It would result in a division by zero error if it was zero.
- limx→ac = c
- limx→axn = an
Special Rules
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Limit of a function of Two Variables
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The 1-dimensional limit of a function evaluated across a curve changes depending on the curve chosen, which is a popular technique to prove that a function of two variables is not continuous at a point.
The limit of a two-variable function demands that f(x,y) be within L whenever (x,y) is within L. (a,b). A two-variable function is continuous at a location if the limit, the function, and the limit and function are identical.
If we have a function f(x,y) that depends on two variables x and y, we can claim that this function has a limit, say, C as (x,y)→(a,b) provided that \(\epsilon\)> 0 , there exists Δ>0 such that |f(x,y)-C|<\(\epsilon\) whenever 0<\(\sqrt{(x - a)^2 + (y -b)^2}\)< Δ.
It defined as lim(x,y)→(a,b)f(x,y) = C
Limit Detailed Video Explanation
Limit of a Function and Continuity
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The function's limits and its continuity are closely linked to each other. Continuous and discontinuous functions are both possible. A two-variable function is continuous if the limit, the function, and the limit and function are identical at that point.
The condition f(X) →λ as x → a in elementary calculus indicates that the number f(x) can be set to lay as near to the number λ, we desire as long as we take the number unequal to a but close enough to a. We call a function continuous if it equals its limit, just as we do with a single variable:
If lim(x,y) → (a,b)f(x,y)=f(a,b) function f(x,y) is continuous at the point (a,b) (a,b).
If a function is continuous on a domain D, it is continuous at all points on the domain.
We employ the following key result in the derivation of functions: f'(a) of a given function f at a number a can be thought of as,
f'(a) = limx→a [f(x)−f(a)]/(x−a)
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Limit of Complex Function
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Follow the formula below to differentiate functions of a complex variable:
\(lim_{\bigtriangleup z \to 0} \frac{ f(z_0 + \bigtriangleup z) - f(z_0)}{\bigtriangleup z}\)
For the above formula, the function f(z) is said to be differentiable at z = z0
Here, Δz = Δx + i Δy
Limits of Exponential Function
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>An exponential function's limit is the same as the limit of an exponent with the same base.
For any real number x, the exponential function f with the base a is f(x) = ax where a >0 and a is not equal to zero. Some of the most essential laws of limits employed while dealing with exponential function limitations are listed below.
For f(b)>1
- limx→∞bx = ∞
- lim x→ -∞bx=0
For 0<b<1
- lim x→∞bx=0
- limx→ -∞bx=∞
Things to Remember
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>>- Limits are the values at which a function approaches the output for the given input values.
- Limits are used to define integrals, derivatives, and continuity, among other things, in calculus and mathematical analysis.
- It is employed in the analysis process and always refers to the function's behaviour at a specific time.
- Some Properties of Limits:
- limx→a[c.f(x)] = c. limx→af(x)
- limx→a[f(x)±g(x)] = limx→af(x) ± limx→ag(x)
- limx→a[f(x)⋅g(x)] = limx→af(x)⋅limx→ag(x)
- limx→a[f(x).g(x)] = limx→af(x) . limx→ag(x)
- limx→a[f(x)/g(x)] = [lim x→af(x)] / [limx→ag(x)] provided limx→ag(x)≠0
- limx→ac = c
- limx→axn = an
- Rules of Limits:
Important Questions Based on Limits
Ques: Evaluate: limx→∞(x2+a2)1/2 - (x2+b2)1/2 / (x2+c2)1/2 - (x2+d2)½ (2 Marks)
Ans:
Ques: To Compute lim x→-45x2+8x-3 (3 Marks)
Ans: Divide the limit into three parts using property 2. Then, using property 1, extract the constants from the first two equations. This results in,
limx→4(5x2+8x-3)=limx→-45x2+limx→-4(8x)-limx→-4(3)
=5(-4)2+8(-4)-3
=80-32-3
=45
Ques: To Compute lim x→6\(\frac{(x-3)(x-2)}{x-4)}\)(3 Marks)
Ans:
Ques: Compute lim x→3(x2-9)/ (x-3) (3 Marks)
Ans: Given, limx→3(x2-9)/(x-3)
It should be noticed that if the value 3 is explicitly substituted into the function, the numerator and denominator will both become 0, and we know that the value 0/0 , does not exist.
Using the property of squares, we have:
= limx→3(x-3)(x+3) / (x-3)
= limx→3(x+3)
=6
Ques: Find lim x→∞ sinx/x. (3 Marks)
Ans:
so that x → ∞ ⇒ y → 0
∴ limx→∞ (sin x / x) = limy→0 (y. sin (1 / y))
=limy→0 y. limy→0 sin (1 / y)
= 0
Ques: If f(a)=2,f'(a)=1,g(a)=-1;g'(a)=2 , then find limx→a \(\frac{g(x)f(a) - g(a)f(x)}{x-a}\)?(3 Marks)
Ans:
Ques: Find the limit limn→∞ [1/n2 + 2/n2 + … + n/n2] (3 Marks)
Ans: limn→∞ [1/n2 + 2/n2 + … + n/n2]
= limn→∞ [1+2+3+…. +n]/ n2
= limn→∞ [(n / 2) * ( n+1)] / n2
= ½ limn→∞ ( n+1) / n
= ½ limn→∞ (1 + 1/n)
= -½
Ques: let f: R→R be such that f (1) = 3 and f’(1) = 6. Then find the value of lim x→0 [f (1+x) / f (1)]1/x. (3 Marks)
Ans: Let y = [f (1 + x) / f (1)]1/x
So, log y = 1/x [log f (1 + x) – log f (1)]
So, limx→0 log y = limx→0 [1 / f (1 + x) . f’(1 + x)]
= f’(1) / f(1)
= 6/3
log (limx→0 y) = 2
limx→0 y = e2
Ques: If f (x) = [2/x ] − 3, g (x) = x − [3 / x] + 4 and h (x) = −[2 (2x + 1)] / [x2 + x − 12], then what is the value of limx→3 [f (x) + g (x) + h (x)]? (3 Marks)
Ans: We have f (x) + g (x) + h (x) = [x2 − 4x + 17− 4x − 2] / [x2 + x − 12]
= [x2 − 8x + 15] / [x2 + x − 12]
= [(x − 3) (x − 5)] / [(x − 3) (x + 4)]
∴limx→3 [f (x) + g (x) + h (x)]
= limx→3 (x − 3) (x − 5) / (x − 3) (x + 4)
= −2/7
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