What is a Limit in calculus? A limit can be explained as the value which a function tends to approach as the input value (also known as index) gains some value. It is the converging of different values at a point. Boundedness of a function is shown by limits. In this article, we come across solved examples of limits.
Consider f(x) to be a function. In a function, if x takes a definite value say b, x → b is called limit. Here ‘b’ is a value which is pre-assigned. It is represented as limx→bf(x).
The tendency of f(x) at x=a towards the left is called left limit and denoted by limx→a– and towards the right is called right limit denoted by limx→a+. Limit of a function at a point is the common value of the right and left hand limits, if they coincide.
The graphical representation of limits is as follows:
Algebra of Limits
Suppose limx→a a(x) = r and limx→a b(x) = t then the following can be defined.
- limx→a [r(x) + t(x)] = limx→a r(x) + limx→a t(x).
- limx→a [r(x) − t(x)] = limx→a r(x) − limx→a t(x).
- limx→a [r(x) × t(x)] = [limx→a r(x)] × [limx→a t(x)].
- limx→a [r(x) ÷ t(x)] = [limx→a r(x)] ÷ [limx→a t(x)].
- limx→a [α.r(x))] = α. limx→a r(x).
Some Standard Limits
The following are some of the standard limits.
2] \ \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\tan x}}{x}} \right] = 1\\
3] \ \mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{x}} \right)}^x} = e\\
4] \ \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\\
5] \ \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_a}\left( {1 + x} \right)}}{x} = \frac{1}{{\ln \,a}}\\
6] \ \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} – 1}}{x} = \ln \,\,a\\
7] \ \mathop {\lim }\limits_{x \to 1} \frac{{{x^m} – 1}}{{x – 1}} = m\\
8] \ \mathop {\lim }\limits_{x \to a} \frac{{{x^m} – {a^m}}}{{x – a}} = m\,\,{a^{m – 1}}\\
9] \ \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ – 1}}x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ – 1}}x}}{x} = 1\\\end{array} \)
Related articles
Limits, Continuity and Differentiability
Solved Examples on Limits for Practice
Below are illustrated some of the questions based on limits asked in JEE previous exams.
Example 1: Find limx→∞ sinx/x.
Solution:
Let x = 1/y or y = 1/x, so that x → ∞ ⇒ y → 0
∴ limx→∞ (sin x / x) = limy→0 (y . sin (1 / y))
=limy→0 y . limy→0 sin (1 / y)
= 0
Example 2: If f(a) = 2, f′(a) = 1, g(a) = −1; g′(a) = 2, then find
Solution:
Consider 
Example 3: Evaluate
Solution:
Example 4: Find the limit limn→∞ [1/n2 + 2/n2 + … + n/n2]
Solution:
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
Example 5: Find limx→0 sin (π cos2x) / x2
Solution:
= π (-1).1.(-1) = π
Example 6: 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.
Solution:
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
Example 7: 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)]?
Solution:
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
Example 8:
Solution:
Related Video
Limits, Continuity and Differentiability – Part 1
Limits, Continuity and Differentiability – Part 2
Limits, Continuity and Differentiability – Important Topics
Limits, Continuity and Differentiability – Important Questions
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