Limits :
Let y = f(x) be a given function defined in the neighborhood of x = a, but not necessarily at the point x = a.
Basically we are interested in examining the behavior (or tendency) of the function f(x) when the distance between the points x and a is small, i.e. |x-a| is small but not zero.
The limiting behavior of the function in the neighborhood of x = a when |x–a| is small is called the limit of the function when x approaches a. Mathematically we write this as limx→a f(x).
Let limx→a f(x) = l . It would simply mean that when we approach the point x = a from the values which are just greater than or just smaller than x= a , f(x) would have a tendency to move closer to the value ‘ l ’ .
This is same as saying, “difference between f(x) and l can be made as small as we feel like by suitably choosing x in the neighborhood of x = a .
Mathematically, we write this as, limx→a f(x) = l , which is equivalent to saying that |f(x)-l| < ε ∀ x such that 0 < |x-a|< δ and ε depends on δ .
Where ε and δ are sufficiently small positive numbers.
It is clear from the above discussion that if we are interested in finding the limit of f(x) at x = a, the first thing that we have to make sure is that f(x) is well defined in the neighborhood of x = a and not necessarily at x = a (that means x = a may or may not be in the domain of f(x)), because we have to examine it’s behavior or tendency in the neighborhood of x = a .
Now the following possibilities may arise:
(a) Left tendency of f(x) is same as it’s right tendency, as shown in the adjacent figure.
That means when we approach x= a from the values which are just less than a , f(x) has a tendency to move towards the value ‘ l ’ (left tendency).
Similarly when we approach x = a from the values which are just greater then a, f(x) has a tendency to move towards the value ‘ l ’ (right tendency).
In this case we say f(x) has limit at x = a i.e. limx→a f(x) = l
(b) When the left tendency is not the same as the right tendency, as shown in the adjacent figure.
Clearly left tendency (l1) is not the same as right tendency ( l2). In this case we say limit of f(x) at x = a will not exist.
(c) When the left tendency and/or the right tendency is not fixed, (as shown in the adjacent figure).
It is clear that in this case, the function has erratic behavior in the neighborhood of x = a, and it will not be possible to talk about the left and the right tendencies of the function in the neighborhood of x = a.
In this case we conclude that the limit of f(x) at x = a will not exist.
From the above mentioned discussion, it is clear that the limit of f(x) at x = a would exist if and only if f(x) is well defined in the neighborhood of x = a (not necessarily at x = a) and has a unique behavior in the neighborhood of x = a.
If any of these conditions are not fulfilled the limit of f(x) at x = a will not exist.
Left Limit & Right Limit :
Let y = f(x) be a given function, and x = a is the point under consideration. Left tendency of f(x) at x = a is called it’s left limit and right tendency is called it’s right limit.
Left tendency (left limit) is denoted by f(a − 0) or f(a −) and right tendency (right limit) is denoted by f( a + 0) or f(a +) and are written as
$ f(a-0) = lim_{h\rightarrow 0 } f(a-h) $
$ f(a+0) = lim_{h\rightarrow 0 } f(a+h) $
where ‘ h ‘ is a small positive number.
NOTE : Thus for the existence of the limit of f(x) at x = a, it is necessary and sufficient that
f(a − 0) = f(a + 0), if these are finite or f(a − 0) and f(a + 0) both should be either + ∞ or − ∞
Frequently used limits:
(i) If p(x) is a polynomial , $ lim_{x \rightarrow a} p(x) = p(a)$
(ii) $ lim_{x \rightarrow 0} \frac{sinx}{x} = 1 $
$ lim_{x \rightarrow 0} \frac{tanx}{x} = 1 $
$ lim_{x \rightarrow 0} cosx = 1 $
(iii) $ lim_{x \rightarrow 0} (1 + x)^{1/x} = e $
(iv) $ lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x = e $
(v) $ lim_{x \rightarrow 0} \frac{e^x -1}{x} = 1 $
(vi) $ lim_{x \rightarrow 0} \frac{a^x -1}{x} = ln(a) , a \in R^+$
(vii) $ lim_{x \rightarrow 0} \frac{(1+x)^n -1}{x} = n $
(viii) $ lim_{x \rightarrow a} \frac{x^m -a^m}{x^n – a^n} = \frac{m}{n}a^{m-n}$
(ix) $ lim_{x \rightarrow 0} \frac{ln(1+x)}{x} = 1 $
(x) $ lim_{x \rightarrow 0} \frac{log_a(1+x)}{x} = log_a e , a > 0 , \ne 1 $
(xi) $ lim_{x \rightarrow 0} \frac{sin^{-1}x}{x} = 1 = \lim_{x \rightarrow 0} \frac{tann^{-1}x}{x} $
If $ lim_{x \rightarrow a} f(x) = 0 $ then the following results would be holding true,
(i) $ lim_{x \rightarrow a} \frac{sin f(x)}{f(x)} = \lim_{x \rightarrow a} \frac{tan f(x)}{f(x)} = \lim_{x \rightarrow a} cosf(x) = 1$
(ii) $ lim_{x \rightarrow a} \frac{sin^{-1} f(x)}{f(x)} = \lim_{x \rightarrow a} \frac{tan^{-1} f(x)}{f(x)} = 1$
(iii) $ lim_{x \rightarrow a} \frac{b^{f(x)}-1}{f(x)} = ln(b) , b > 0 $
(iv) $ lim_{x \rightarrow a} (1+ f(x) )^{1/f(x)} = e $
Frequently used Series Expansions :
Following are some of the frequently used series expansions:
$ sinx = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + …. $
$ cosx = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + …. $
$ tanx = x + \frac{x^3}{3!} + \frac{2}{15}x^5 + …. $
$sin^{-1}x = x + \frac{1^2}{3!}x^3 + \frac{1^2 .3^2}{5!}x^5 + \frac{1^2 .3^2 .5^2}{7!}x^7 + ….. $
$tan^{-1}x = x – \frac{x^3}{3} + \frac{x^5}{5} – ….$
$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + …. $
$ a^x = 1 + x . lna + \frac{x^2}{2!} (lna )^2 + …. $ ; a ∈ R+
$(1+x)^n = 1 + n x + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + …. $ ; n ∈ R. |x|< 1 , n is any real number
Illustration : Evaluate the following limits, if exist
(i) $ lim_{x\rightarrow 0 } |x|^{sinx} $
(ii) $ lim_{x\rightarrow 0 } (1+sinx)^{1/x^2} $
(iii) $ lim_{x\rightarrow \infty } (\frac{x^2 + 2x -1}{2x^2 -3x-2})^{\frac{2x+1}{2x-1}} $
(iv) $ lim_{x\rightarrow 1 } (1 + sin\pi x)^{cot \pi x} $
(v) $ lim_{x\rightarrow \infty } \frac{\sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}}} $
(vi) $ lim_{x\rightarrow 0 } \frac{e^{a/x}- e^{-a/x}}{e^{a/x} + e^{-a/x}} $ , a > 0
(vii) $ lim_{x\rightarrow 0 } \frac{tan^{-1}x – sin^{-1}x}{sin^3 x} $
(viii) $ lim_{x\rightarrow 0 } \frac{|x|^\alpha}{e^x} $ ; α ∈ R+
(ix) $ lim_{x\rightarrow \infty} \frac{2 x^{1/2}+ 3 x^{1/3}+ 4 x^{1/4}+ …. + n x^{1/n}}{(3x-4)^{1/2} + (3x-4)^{1/3} + (3x-4)^{1/4} + …. + (3x-4)^{1/n}} $
(x) $lim_{n \rightarrow \infty} n^{-n^2} ((n+1)(n+\frac{1}{2}) (n+\frac{1}{2^2}) ……..(n+\frac{1}{2^{n-1}}))^n $
Solution :
(i) $ lim_{x\rightarrow 0 } |x|^{sinx} $
$\large = lim_{x\rightarrow 0 } e^{ln(|x|)^{sinx}} $
$ \large = lim_{x\rightarrow 0 } e^{sinx ln|x|} $
$ \large = e^{lim_{x\rightarrow 0 }\frac{ln|x|}{cosec x}} $
$\large = e^{lim_{x\rightarrow 0 }\frac{1/x}{- cosec x . cot x}} $
$ \large = e^{lim_{x\rightarrow 0 }\frac{- sin^2 x}{x . cosx}} $
$ \large = e^{lim_{x\rightarrow 0 }-(\frac{sin x}{x})^2 \frac{x}{cosx}} $
$ = e^0 = 1 $
(ii) $ lim_{x\rightarrow 0 } (1 + sinx)^{1/x^2} $ ; (1∞ form)
The given limit does not exist.
(iii) $ lim_{x\rightarrow \infty} (\frac{x^2 + 2x -1}{2x^2 -3x -2})^{\frac{2x+1}{2x-1}} $
$ lim_{x\rightarrow \infty} (\frac{1 + 2/x -1/x^2}{2 -3/x -2/x^2})^{\frac{2+1/x}{2-1/x}} $
$ = (\frac{1+ 0 -0}{2-0-0})^{\frac{2+0}{2-0}} $
$ = \frac{1}{2} $
(iv) $ lim_{x\rightarrow 1 } (1 + sin\pi x)^{cot \pi x} $
$\large = lim_{x\rightarrow 1 } [(1 + sin\pi x)^{\frac{1}{sin\pi x}}]^{cos\pi x} $
= e-1 = 1/e
(v) $ lim_{x\rightarrow \infty } \frac{\sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}}} $
$ = lim_{x\rightarrow \infty } \frac{1}{\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}}}$
$ = \frac{1}{\sqrt{1+0}} = 1 $
(vi) $ L.H.L = \lim_{h\rightarrow 0 } \frac{e^{a/0-h}- e^{-a/0-h}}{e^{a/0-h} + e^{-a/0-h}} $
$ = lim_{h \rightarrow 0 } \frac{e^{-a/h}- e^{a/h}}{e^{-a/h} + e^{a/h}} $
$ = lim_{h \rightarrow 0 } \frac{e^{-2a/h} – 1}{e^{-2a/h} + 1} = -1 $
$ R.H.L = \lim_{h\rightarrow 0 } \frac{e^{a/0+h}- e^{-a/0+h}}{e^{a/0+h} + e^{-a/0+h}} $
$ = lim_{h \rightarrow 0 } \frac{e^{a/h} – e^{-a/h}}{e^{a/h} + e^{-a/h}} $
$ = lim_{h \rightarrow 0 } \frac{1 – e^{-2a/h}}{1 + e^{-2a/h} } = 1 \; \ne L.H.L $
(vii) $ lim_{x\rightarrow 0 } \frac{tan^{-1}x – sin^{-1}x}{sin^3 x} $
$ = lim_{x\rightarrow 0 } \frac{tan^{-1}x – sin^{-1}x}{x^3 (\frac{sin^3 x}{x^3})} $
$ = lim_{x\rightarrow 0 } \frac{tan^{-1}x – sin^{-1}x}{x^3 } $
$ = lim_{x\rightarrow 0 } \frac{tan^{-1}x – tan^{-1} \frac{x}{\sqrt{1-x^2}}}{x^3 } $
(viii) $ lim_{x\rightarrow 0 } \frac{|x|^\alpha}{e^x} $
As , $lim_{x\rightarrow 0 } |x|^{\alpha} = 0 $ and $lim_{x\rightarrow 0 } e^x = 1$
$ lim_{x\rightarrow 0 } \frac{|x|^\alpha}{e^x} = 0 $
(ix) $ lim_{x\rightarrow \infty} \frac{2 x^{1/2}+ 3 x^{1/3}+ 4 x^{1/4}+ …. + n x^{1/n}}{(3x-4)^{1/2} + (3x-4)^{1/3} + (3x-4)^{1/4} + …. + (3x-4)^{1/n}} $
In these type of questions divide Nr and Dr by the highest power of x. (x1/2 in this question).
$ lim_{x\rightarrow \infty} \frac{2 + 3 x^{1/3 – 1/2}+ …. + n x^{1/n – 1/2}}{(3-4/x)^{1/2} + x^{1/3 -1/2}(3-4/x)^{1/3} + …. + x^{1/n – 1/2}(3-4/x)^{1/n}} $
$= \frac{2}{\sqrt{3}}$
Also Read
| →Limits of a Function →Continuity of a Function →Differentiability of a Function →Rules for differentiation →Higher order derivatives →L’HOSPITAL’S RULE |