Why every point of a function where differentiation exists has only one tangent? Can anyone help me out?
Why every point of a function where differentiation exists has only one tangent?
I know the slope at any point of any function is defined by differentiation at that point.But there may be another straight line which touches the function at that point but slope is different from the differentiation value at that point. But this does not happen(I am not considering singular point).

It's not just about a line touching. (Every line through the point touches the graph there.) It's the best linear approximation, in the sense that the error $f(x)-L(x)$ goes to $0$ faster than $x$ approaches $a$. (Here $L(x)$ is the linear function whose graph we're discussing.)
EDIT: To be more precise, if $f$ is differentiable at $x=a$, then the linear function $L(x)=f(a)+f'(a)(x-a)$ is the unique linear function with the property that 
$$\lim_{x\to a}\frac{f(x)-L(x)}{x-a} = 0.$$ 
Note that if we take $L(x)=f(a)+m(x-a)$, this limit becomes
$$\lim_{x\to a}\frac{f(x)-f(a)-m(x-a)}{x-a} = \lim_{x\to a}\frac{f(x)-f(a)}{x-a} -m = f'(a)-m,$$
and this is $0$ if and only if $m=f'(a)$. 
If $f$ fails to be differentiable, this limit will never be 0. Try it, for example, with $f(x)=|x|$ at $x=0$. If you take a line of slope $m$ and $x>0$, this limit is $\lim_{x\to 0^+} \dfrac{x-mx}x = 1-m$, so $m$ would have to be $1$, but if $x<0$, this limit is $\lim_{x\to 0^-} \dfrac{-x-mx}x = -1-m$, so $m$ would have to be $-1$. There therefore is no tangent line at $x=0$.
