Question

find the value of k in each function F is continuous at the
indicated point


x not equals to 0
X equals to 0​

Answer 1

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

Given that,

$\:\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: f(x)-\begin{cases} &\sf{\dfrac{x}{ |x| + {2x}^{2}} \: \: when \: \: x \: \ne \: 0 } \\ &\sf{k \: \: when \: \: x = 0} \end{cases}\end{gathered}\end{gathered}$

is continuous at x = 0.

We know,

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

$\bf :\longmapsto\:\tt \:\lim_{x\to \: a^-} = \tt \:\lim_{x\to \: a^ + } = f(a)$

Now,

Step :- 1

$\bf :\longmapsto\:f(0) = k - - - (1)$

Step :- 2 Evaluation of Left Hand Limit

$\rm :\longmapsto\:\tt \:\lim_{x\to \: 0^-} \: \dfrac{x}{ |x| + {2x}^{2} }$

$\red{\rm :\longmapsto\:Put \: x = 0 \: - \: h = - \: h} \\ \red{\rm :\longmapsto\:as \: x \to \: 0 \: so \: h \to \: 0 \: \: \: \: \: }$

$\rm \: \: = \: \tt \:\lim_{h\to \: 0} \: \dfrac{ - h}{ | - h| + 2 {( - h)}^{2} }$

$\rm \: \: = \: \tt \:\lim_{h\to \: 0} \: \dfrac{ - h}{h+ 2 {h}^{2} }$

$\rm \: \: = \: \tt \:\lim_{h\to \: 0} \: \dfrac{ - h}{h(1+ 2 {h})}$

$\rm \: \: = \: \tt \:\lim_{h\to \: 0} \: \dfrac{ - 1}{(1+ 2 {h})}$

$\rm \: \: = \: - 1$

Hence,

$\rm :\longmapsto\:\tt \:\lim_{x\to \: 0^-} \: \dfrac{x}{ |x| + {2x}^{2}} = - 1 - - - (1)$

Step :- 3 Evaluation of Right Hand Limit

$\rm :\longmapsto\:\tt \:\lim_{x\to \: 0^ + } \: \dfrac{x}{ |x| + {2x}^{2} }$

$\red{\rm :\longmapsto\:Put \: x = 0 \: + \: h = \: h} \\ \red{\rm :\longmapsto\:as \: x \to \: 0 \: so \: h \to \: 0 \: \: \: \: \: }$

$\rm \: \: = \: \tt \:\lim_{h\to \: 0} \: \dfrac{h}{ |h| + 2 {(h)}^{2} }$

$\rm \: \: = \: \tt \:\lim_{h\to \: 0} \: \dfrac{h}{h+ 2 {h}^{2} }$

$\rm \: \: = \: \tt \:\lim_{h\to \: 0} \: \dfrac{h}{h(1+ 2 {h})}$

$\rm \: \: = \: \tt \:\lim_{h\to \: 0} \: \dfrac{1}{(1+ 2 {h})}$

$\rm \: \: = \: 1$

Hence,

$\rm :\longmapsto\:\tt \:\lim_{x\to \: 0^ + } \: \dfrac{x}{ |x| + {2x}^{2} } = 1 - - - (2)$

From, equation (1) and (2), we concluded that

$\rm :\longmapsto\:\tt \:\lim_{x\to \: 0^-} \: \dfrac{x}{ |x| + {2x}^{2}} \: \ne \: \:\tt \:\lim_{x\to \: 0^ + } \: \dfrac{x}{ |x| + {2x}^{2}}$

So, f(x) can never be Continuous.

So, k can take any real value.

Additional Information :-

If f and g are two Continuous functions then

• 1. f + g is continuous.

• 2. f - g is continuous.

• 3. f.g is continuous.

Every differentiable function is always continuous but every continuous function may or may not be differentiable.

Every sine, cosine, exponential and Logarithmic functions are always continuous and differentiable on its domain.