in some cases we take lhl and rhl and in some we dont why is that so when - Maths - Continuity and Differentiability

Question

in some cases we take lhl and rhl and in some we dont why is
that so? when do we take lhl and rhl

Ankush Jain · Accepted Answer

Those functions which are defined in such a way that they take different values just before and just after the point where you are discussing the limit of the function, then we find L.H.L. and R.H.L. shown below in the example :

$f (x) = {\begin{matrix} \frac{x + 2}{x + 5} if x < 2 \\ x + 3 if x \geq 2 \end{matrix}$

Now, when we take L.H.L., then the value of f (x) will be $\frac{x + 2}{x + 5}$ as x is slightly less than 2.

$So, L.H.L. = \lim_{x \to 2^{-}} \frac{x + 2}{x + 5} = \lim_{h \to 0} \frac{2 - h + 2}{2 - h + 5} = \frac{4}{7} So, R.H.L. = \lim_{x \to 2^{+}} (x + 3) = \lim_{h \to 0} (2 + h + 3) = 5 + 0 = 5$

Since L.H.L. ≠ R.H.L.

So, limit does not exists.

If the function is defined so that the values just before or just after the point say 'a' at which we are finding the limit are same, then we find only the $\lim_{x \to a} f (x)$ .

For example –

${\begin{matrix} - x + 3 if x ≢ 3 \\ 0 if x = 3 \end{matrix}$

Here, if x is slightly less than 3 or slightly more than 3, in each case we get x ≠ 3.

So function will be –x + 3.

$∴ \lim_{x \to 3} f (x) = \lim_{x \to 3} (- x + 3) = 0$

and also f (3) = 0

$∴ f (3) = \lim_{x \to 3} f (x)$

So limit exists.

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in some cases we take lhl and rhl and in some we dont. why is that so? when do we take lhl and rhl