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Question: Find \[\displaystyle \lim_{x\rightarrow 0} f\left( x\right)\] and \[\displaystyle \lim_{x\rightarrow...

Find lim⁡x→0f(x)\displaystyle \lim_{x\rightarrow 0} f\left( x\right) and lim⁡x→1f(x)\displaystyle \lim_{x\rightarrow 1} f\left( x\right) where f(x)={2x+3,x;≤0 3(x+1),x;>0f\left( x\right) =\begin{cases}2x+3,&x;\leq 0\\\ 3\left( x+1\right) ,&x;>0\end{cases}.

Explanation

Solution

Hint: In this question it is given that we have to find the value of lim⁡x→0f(x)\displaystyle \lim_{x\rightarrow 0} f\left( x\right) and lim⁡x→1f(x)\displaystyle \lim_{x\rightarrow 1} f\left( x\right) . So for finding the limits of a function in a particular point (let the point be a) we have to use one formula that is, lim⁡x→af(x)=lim⁡x→a−f(x)=lim⁡x→a+f(x)\displaystyle \lim_{x\rightarrow a} f\left( x\right) =\displaystyle \lim_{x\rightarrow a^{-}} f\left( x\right) =\displaystyle \lim_{x\rightarrow a^{+}} f\left( x\right) =L. So according to this formula we can say that if the left hand limit(i.e, LHL) and the right hand limit(i.e, RHL) is equal and if their value is ‘L’ then the limit exists and the value of the limit is equal to L.

Complete step-by-step answer:
First of all we have to find the limit of f(x) at a particular point x=0,
For this,
LHL,
lim⁡x→0−f(x)\displaystyle \lim_{x\rightarrow 0^{-}} f\left( x\right)
=lim⁡x→0(2x+3)=\displaystyle \lim_{x\rightarrow 0} \left( 2x+3\right)
=(2×0+3)=\displaystyle \left( 2\times 0+3\right) =3,
RHL,
lim⁡x→0+f(x)\displaystyle \lim_{x\rightarrow 0^{+}} f\left( x\right)
=lim⁡x→03(x+1)=\displaystyle \lim_{x\rightarrow 0} 3\left( x+1\right)
=3(0+1)=3\left( 0+1\right) = 3,
Therefore, we can say that LHL=RHL=3.
So the value of lim⁡x→0f(x)\displaystyle \lim_{x\rightarrow 0} f\left( x\right) is 3.

Now for the second part of the question, that is the limit of f(x) at a particular point x=1, so for this,
LHL,
lim⁡x→1−f(x)\displaystyle \lim_{x\rightarrow 1^{-}} f\left( x\right)
=lim⁡x→13(x+1)=\displaystyle \lim_{x\rightarrow 1} 3\left( x+1\right)
= 3(1+1) = 3×2 3\times 2=6,

RHL,
lim⁡x→1+f(x)\displaystyle \lim_{x\rightarrow 1^{+}} f\left( x\right)
=lim⁡x→13(x+1)=\displaystyle \lim_{x\rightarrow 1} 3\left( x+1\right)
=3(1+1)=3\left(1+1\right) = 3×2 3\times 2=6,
Therefore, we can say that LHL=RHL=6.
So the value of lim⁡x→1f(x)\displaystyle \lim_{x\rightarrow 1} f\left( x\right) is 6.

So our required solutions are
lim⁡x→0f(x)\displaystyle \lim_{x\rightarrow 0} f\left( x\right) =3,
lim⁡x→1f(x)\displaystyle \lim_{x\rightarrow 1} f\left( x\right) = 6.

Note: Whenever we face such types of problems first find the LHL and then RHL for a particular point and if they equal then the limit of the given function exists. Also while applying LHL in any particular point(say point a) x→a−x\rightarrow a^{-} means x<ax < a, and according to that we have to put the value of function for a piecewise function.