Question

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Answer 1

$\\\\\huge{⚝}$ $\\\\\huge{\underline{\mathcal{Answer}}}$

$\\\huge{☞}$$\\\sf{k = 2}$

$\\\\\huge{⚝}$ $\\\\\huge{\underline{\mathcal{Given:}}}$

$f(x) = \begin{cases} \dfrac{\sin 5x}{x^2+2x} & \text{, } x \neq 0 \\ k+\dfrac{1}{2} & \text{, x = 0}\end{cases}$

is continuous at x = 0.

$\\\\\huge{⚝}$ $\\\\\huge{\underline{\mathcal{To Find:}}}$

$\huge{→}$Value of k

$\\\\\huge{⚝}$ $\\\\\huge{\underline{\mathcal{Steps}}}$

Here, f(0) = $k + \dfrac{1}{2}$

First we will calculate for $0^{-}$:

$\\\large{⇒}$$\\\lim_{x\to {0}^{ - } }f(x) = \lim_{x\to {0}^{ - } } \: \dfrac{\sin \: 5x}{ {x}^{2} + 2x }$

$\\\large{⇒}$$\\ \lim_{x\to {0}^{ - } }f(x) =\lim_{x\to {0}^{ - } } \dfrac{5 \times \sin5x}{5 \times x(x + 2)}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ - } }f(x) = \lim_{x\to {0}^{ - } } \dfrac{5}{x + 2} \times \lim_{x\to {0}^{ - } } \dfrac{ \sin5x }{5x}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ - } }f(x) = \lim_{x\to {0}^{ - } } \dfrac{5}{x + 2} \times 1$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ - } }f(x) =\lim_{x\to {0}^{ - } } \dfrac{5}{x + 2}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ - } }f(x) = \dfrac{5}{0 + 2}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ -} }f(x) = \dfrac{5}{2}$

Now we will calculate for $0^{+}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ + } }f(x) = \lim_{x\to {0}^{ + } } \: \dfrac{\sin \: 5x}{ {x}^{2} + 2x }$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ + } }f(x) =\lim_{x\to {0}^{ + } } \dfrac{5 \times \sin5x}{5 \times x(x + 2)}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ + } }f(x) = \lim_{x\to {0}^{ + } } \dfrac{5}{x + 2} \times \lim_{x\to {0}^{ + } } \dfrac{ \sin5x }{5x}$

$\\\large{⇒}$$\\ \lim_{x\to {0}^{ + } }f(x) = \lim_{x\to {0}^{ + } } \dfrac{5}{x + 2} \times 1$

$\\\large{⇒}$$\\ \lim_{x\to {0}^{ + } }f(x) = \lim_{x \to {0}^{ + } } \dfrac{5}{x + 2}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ + } }f(x) = \dfrac{5}{0 + 2}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ + } }f(x) = \dfrac{5}{2}$

For f(x) to be continuous at x = 0,

$\large{⇒}$$\lim_{x\to {0}^{ - } } = \lim_{x\to {0}^{ + } } = \lim_{x\to 0}$

$\\\large{⇒}$$\\\lim_{x\to {0}^{ - } } = \lim_{x\to {0}^{ + } } = \dfrac{5}{ 2}$

$\\\large{⇒}$$\\ \lim_{x\to {0}^{ - } } = \lim_{x\to{0}}$

$\\\large{⇒}$$\\ \lim_{x\to {0}^{ - } } = k + \dfrac{1}{2}$

$\\\large{⇒}$$\\ \dfrac{5}{2} = k + \dfrac{1}{2}$

$\\\large{⇒}$$\\ \dfrac{5}{2} - \dfrac{1}{2} = k$

$\\\large{⇒}$$\\ k = \dfrac{4}{2}$

$\large{⇒}$$k = 2$

$\\\\\fbox{\fbox{\fbox{\fbox{\huge{\underline{\underline{\sf{\green{Hope\:it\:helps}}}}}}}}}$

Answer 2

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

Given that

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{\dfrac{sin5x}{ {x}^{2} + 2x} , \: \: \: x \: \ne \: 0} \\ \\ &\sf{k \: + \: \dfrac{1}{2} , \: \: \: x \: = \: 0} \end{cases}\end{gathered}\end{gathered}$

is continuous at x = 0.

Consider,

$\rm :\longmapsto\:f(0) = k + \dfrac{1}{2}$

Now,

Consider,

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \tt f(x)}$

$\rm \: = \: \: \displaystyle\lim_{x \to 0} \tt \dfrac{sin5x}{ {x}^{2} + 2x}$

If we substitute directly x = 0, we get indeterminant form.

So,

can be further rewritten as

$\rm \: = \: \: \displaystyle\lim_{x \to 0} \tt \dfrac{sin5x}{ x \: ({x} + 2)}$

can be rewritten as

$\rm \: = \: \: \displaystyle\lim_{x \to 0} \tt \dfrac{1}{x + 2} \times \displaystyle\lim_{x \to 0} \tt \dfrac{sin5x}{x}$

$\rm \: = \: \: \tt \dfrac{1}{0+ 2} \times \displaystyle\lim_{x \to 0} \tt \dfrac{sin5x}{5x} \times 5$

We know,

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

So, using this, we get

$\rm \: = \: \: \dfrac{1}{2} \times 1 \times 5$

$\rm \: = \: \: \dfrac{5}{2}$

So,

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \tt f(x) \: = \: \dfrac{5}{2} }$

Now,

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

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to a} \tt f(x) = f(a)}$

Now, According to statement, it is given that function f(x) is continuous at x = 0.

So,

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \tt f(x) = f(0)}$

$\rm :\longmapsto\:\dfrac{5}{2} = k + \dfrac{1}{2}$

$\rm :\longmapsto\:k \: = \: \dfrac{5}{2} - \dfrac{1}{2}$

$\rm :\longmapsto\:k \: = \: \dfrac{5 - 1}{2}$

$\rm :\longmapsto\:k \: = \: \dfrac{4}{2}$

$\bf\implies \:k \: = \: 2$

• So, Option (1) is correct

Additional Information :-

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

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

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

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

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

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{sin^{ - 1} x}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{tan^{ - 1} x}{x} = 1}}$