A given function f(x) can be discontinuous at x = a due to following reasons

(1) f(x) is not defined at x = a, i.e f(a) not defined.

or (2) $$\mathop {\lim }\limits_{x \to a} f\left( x \right)$$ does not exists,

or (3) $$f\left( a \right) \ne \mathop {\lim }\limits_{x \to a} f\left( x \right)$$

If I represent a integer then

$\underset{x\to I^{+}}{\lim }\left[x\right]=I$

and $\underset{x\to I^{-}}{\lim }\left[x\right]=I-1$

Also

$\underset{x\to I^{+}}{\lim }\left\{x\right\}=0$

and $\underset{x\to I^{-}}{\lim }\left\{x\right\}=1$

$\underset{x\to a}{\lim }f(x)$ is said to exists if when x is approaching to "a" (from both the sides) then the value of the function f(x) is approaching to same finite value.

Thus, $\underset{x\to a}{\lim }f(x)$ exists $\iff$. $\underset{x\to a^{-}}{\lim }f(x)=\underset{x\to a^{+}}{\lim }f(x)$ = L (L must be finite)

Here, $\underset{x\to a^{-}}{\lim }f(x)$ = Left hand limit of function f(x) at x = a, it means when you are approaching towards "a" from left side of "a" then what is the value of the function f(x).

and $\underset{x\to a^{+}}{\lim }f(x)$ = Right hand limit of function f(x) at x = a, it means when you are approaching towards "a" from right side of "a" then what is the value of the function f(x).

A function f(x) is said to be continuous at x = a if and ony if

(1) f(x) is defined at x = a, i.e f(a) defined.

and (2) $\underset{x\to a}{\lim }f\left(x\right)$ exist,

and (3) $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$

$\therefore$ We can say function f(x) is said to be continuous at x = a when

$f\left(a\right)=\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }f\left(x\right)$ = Finite

If $\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }f\left(x\right)$ = M then

limiting value of the function f(x) or $\underset{x\to a}{\lim }f\left(x\right)=$ M

Note : Here M may or may not be finite.

If the function f(x) is continuous at x = a then

$\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$

provided that x = a should not be the boundary point of f(x).

Note : All polynomial, logarithamic, exponential, trigonometric and inverse trigonometric functions are continuous in their domain.

For solving this limit $\lim \limits_{x \rightarrow a} f(x)$ sometime we put $x={a+h \text { }}$ or $x={a-h \text { }}$ and we take $$\mathop {\lim }\limits_{h \to 0} $$

$ \text { i.e } \lim \limits_{x \rightarrow a} f(x) =\lim \limits_{h \rightarrow 0} f(a+h)= \\ \lim \limits_{h \rightarrow 0} f(a-h) $

Note : (1) $\lim \limits_{x \rightarrow a^{+}} f(\underline{x})=\lim \limits_{h \rightarrow 0^{+}} f(a+h)=\lim \limits_{h \rightarrow 0^{-}} f(a-h)$

(2) $\lim \limits_{x \rightarrow a^{-}} f({x})=\lim \limits_{h \rightarrow 0^{+}} f(a-h)=\lim \limits_{h \rightarrow 0^{-}} f(a+h)$

This rule states that, if $\lim \limits_{x \rightarrow a} \frac{f(x)}{g(x)}$, reduces to $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Then, differentiate numerator and denominator till this form is removed.

i.e. $\lim \limits_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim \limits_{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$, provided the later limit exists.

But, if it again take form $\left(\frac{0}{0}\right.$ or $\left.\frac{\infty}{\infty}\right)$,

then $\lim \limits_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim \limits_{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lim \limits_{x \rightarrow a} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)}$

and this process is continued till $\left(\frac{0}{0}\right.$ or $\left.\frac{\infty}{\infty}\right)$ form is removed.

Note :

(1) L'Hospital's rule is applicable to only two indeterminate forms $\left(\frac{0}{0}\right.$ or $\left.\frac{\infty}{\infty}\right)$.

(2) f(x) and g(x) should be differentiable at x = a and f'(x) and g'(x) must be continuous at x = a then you can apply L'Hospital's Rule.

Sometimes, following expansions are useful in evaluating limits. You should remember those expansions :

1. $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5} \cdots(-1 < x \leq 1)$

2. $\log (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5} \cdots(-1 < x \leq 1)$

3. $e^x=1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\frac{x^4}{4 !} \cdots$

4. $e^{-x}=1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\frac{x^4}{4 !} \cdots$

5. $a^x=1+x\left(\log _e a\right)+\frac{x^2}{2 !}\left(\log _e a\right)^2+\cdots$

6. $\sin x=x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\frac{x^7}{7 !} \cdots$

7. $\cos x=1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !} \cdots$

8. $\tan x=x+\frac{x^3}{3}+\frac{2}{15} x^5+\cdots$

9. $(1+x)^n=1+\frac{n x}{1 !}+\frac{n(n-1) x^2}{2 !}$ $$ +\frac{n(n-1)(n-2) x^3}{3 !}+\ldots n \in R \text { and }|x|<1 $$

10. $\frac{x^n-a^n}{x-a}=x^{n-1}+x^{n-2} a+x^{n-3} a^2+\ldots+a^{n-1}$

11. $(1+x)^{1 / x}=e\left(1-\frac{x}{2}+\frac{11 x^2}{24}+\ldots\right)$

12. $\sin ^{-1} x=x+\frac{1^2}{3 !} x^3+\frac{1^2 \cdot 3^2}{5 !} x^5+\frac{1^2 \cdot 3^2 \cdot 5^2}{7 !} x^7+\ldots$

13. $\tan ^{-1} x=x-\frac{x^3}{3}+\frac{x^5}{5}+\ldots$

14. $\sec ^{-1} x=1+\frac{x^2}{2 !}+\frac{5 x^4}{4 !}+\frac{61 x^6}{6 !}+\ldots$

14. $\left(\sin ^{-1} x\right)^2=\frac{2}{2 !} x^2+\frac{2 \cdot 2^2}{4 !} x^4+\frac{2 \cdot 2^2 \cdot 4^2}{6 !} x^6+\ldots$

15. $x \cot x=1-\frac{x^3}{3}+\frac{x^4}{45}-\frac{2 x^6}{945}+\ldots$

16. $\sec x=1+\frac{x^2}{2}+\frac{5 x^4}{24}+\frac{61 x^6}{720}+\ldots$

17. $x \operatorname{cosec} x=1+\frac{x^2}{6}+\frac{7 x^4}{360}+\frac{31 x^6}{15120}+\ldots$

If direct substitution of $x=a$ while evaluating $\underset{x\to a}{\lim }(x)$ leads to one of the following forms

$\frac{0}{0},\frac{\infty }{\infty },\infty -\infty ,1^{\infty },0^0,{\infty }^0,\infty \times 0$

then it is called indeterminate form.

Note : (1) Here 0, 1 and $\infty$ are not exact values.

(2) Limit of $\infty -\infty ,1^{\infty },0^0,{\infty }^0,\infty \times 0$ can't be find directly, only $\frac{0}{0},\frac{\infty }{\infty }$ can be find directly. That is why if you get any one of those forms $\infty -\infty ,1^{\infty },0^0,{\infty }^0,\infty \times 0$, first convert into $\frac{0}{0}or\frac{\infty }{\infty }$ form and then find limit.

Let $\lim \limits_{x \rightarrow a} f(x)=$ and $\lim\limits _{x \rightarrow a} g(x)=m$. If $\ell$ and $m$ are finite, then:

(A) $\quad \lim \limits_{x \rightarrow a}\{f(x) \pm g(x)\}=\ell \pm m$

(B) $\quad \lim \limits_{x \rightarrow a}\{f(x) \cdot g(x)\}=\ell \cdot m$

(C) $\quad \lim \limits_{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\ell}{m}$, provided $m \neq 0$

(D) $\quad \lim \limits_{x \rightarrow a} k f(x)=k \lim \limits_{x \rightarrow a} f(x)=k \ell$; where $k$ is a constant.

(E) $\quad \lim \limits_{x \rightarrow a} f(g(x))=f\left(\lim \limits_{x \rightarrow a} g(x)\right)=f(m)$; provided $f$ is continuous at $g(x)=m$.

(a) (i) $\quad \lim \limits_{x \rightarrow 0} \frac{\sin x}{x}=\lim \limits_{x \rightarrow 0} \frac{\tan x}{x}=1$ [Where $x$ is measured in radians ]

(ii) $\quad \lim \limits_{x \rightarrow 0} \frac{\tan ^{-1} x}{x}=\lim \limits_{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=1$

(iii) $\quad \lim \limits_{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e \quad ; \quad \lim \limits_{x \rightarrow 0}(1+a x)^{\frac{1}{x}}=e^a$

(iv) $\quad \lim \limits_{x \rightarrow 0}\left(1+\frac{1}{x}\right)^x=$ e $; \lim \limits_{x \rightarrow 0} \left(1+\frac{a}{x}\right)^x=e^a$

(v) $\quad \lim \limits_{x \rightarrow 0} \frac{e^x-1}{x}=1 \quad ; \lim \limits_{x \rightarrow 0} \frac{a^x-1}{x}=\log _e a=\ell$ na $\quad, a>0$

(vi) $\quad \lim \limits_{x \rightarrow 0} \frac{\ell n(1+x)}{x}=1$

(vii) $\quad \lim \limits_{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}$

(b) If $$\mathop {\lim }\limits_{x \to a} f\left( x \right) = 0$$, when $x \rightarrow a$, then

(i) $\quad \lim \limits_{x \rightarrow a} \frac{\sin f(x)}{f(x)}=1$

(ii) $\lim \limits_{x \rightarrow a} \cos f(x)=1$

(iii) $\quad \lim \limits_{x \rightarrow a} \frac{\tan f(x)}{f(x)}=1$

(iv) $\lim \limits_{x \rightarrow a} \frac{e^{f(x)}-1}{f(x)}=1$

(v) $\quad \lim \limits_{x \rightarrow a} \frac{b^{f(x)}-1}{f(x)}=\ell n b, \quad(b>0)$

(vi) $\lim \limits_{x \rightarrow a} \frac{\ell n(1+f(x))}{f(x)}=1$

(vii) $\quad \lim \limits_{x \rightarrow a}(1+f(x))^{\frac{1}{f(x)}}=e$

(viii) $$\mathop {\lim }\limits_{x \to a} {{1 - \cos \left( {f\left( x \right)} \right)} \over {{{\left( {f\left( x \right)} \right)}^2}}} = {1 \over 2}$$

(c) $\quad \lim \limits_{x \rightarrow a} f(x)=A>0$ and $\lim \limits_{x \rightarrow a} \phi(x)=B\left(a\right.$ finite quantity), then $\lim \limits_{x \rightarrow a}[f(x)]^{](x)}=A^B$.