Question

lamda =? when x=2 is continuous​

Answer 1

-2 will be the answer.

As it is continuous at x=2 left hand limit should be equal to right hand limit.

Answer 2

Question :-

Find the value of lambda if f(x) is continuous at x = 2.

$\rm \: \begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{3x - 4, \: \: 0 \leqslant x \leqslant 2} \\ \\ &\sf{2x + \lambda , \: \: 2 < x \leqslant 3} \end{cases}\end{gathered}\end{gathered}$

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: \begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{3x - 4, \: \: 0 \leqslant x \leqslant 2} \\ \\ &\sf{2x + \lambda , \: \: 2 < x \leqslant 3} \end{cases}\end{gathered}\end{gathered}$

Consider,

$\rm \: f(2) = 3(2) - 4 = 6 - 4 = 2$

Consider, RHL

$\rm \: \displaystyle\lim_{x \to 2^ + }\rm f(x)$

$\rm \: = \: \displaystyle\lim_{x \to 2^ + }\rm 2x + \lambda$

To evaluate this limit, we use method of Substitution.

So, Substitute

$\rm \: x \: = \: 2 + h, \: \: as \: x \: \to \: 2, \: \: so \: \: h \: \to \: 0$

So, on substituting these values, we get

$\rm \: = \: \displaystyle\lim_{h \to 0 }\rm [2(2 + h) + \lambda ]$

$\rm \: = \: 4 + \lambda$

We know,

A function f(x) is said to be continuous at x = a, iff

$\rm \: \: \: f(a) \: = \: \displaystyle\lim_{x \to a^-}\rm f(x) \: = \: \displaystyle\lim_{x \to a^ + }\rm f(x)$

So, on substituting the values, we get

$\rm \: 4 + \lambda = 2$

$\rm \: \lambda = 2 - 4$

$\rm \: \lambda \: = \: - \: 2$

Hence,

$\rm\implies \: \: \boxed{\sf{ \:\rm \: \: \lambda \: = \: - \: 2 \: \: }}$

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Additional Information :-

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \: \frac{sinx}{x} \: = \: 1 \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \: \frac{tanx}{x} \: = \: 1 \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \: \frac{log(1 + x)}{x} \: = \: 1 \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{x} - 1}{x} \: = \: 1 \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \: \frac{ {a}^{x} - 1}{x} \: = \: loga \: }}$