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[tex]\\\\\huge{⚝}[/tex] [tex]\\\\\huge{\underline{\mathcal{Answer}}}[/tex]
[tex]\\\huge{☞}[/tex][tex]\\\sf{k = 2}[/tex]
[tex]\\\\\huge{⚝}[/tex] [tex]\\\\\huge{\underline{\mathcal{Given:}}}[/tex]
[tex]f(x) = \begin{cases} \dfrac{\sin 5x}{x^2+2x} & \text{, } x \neq 0 \\ k+\dfrac{1}{2} & \text{, x = 0}\end{cases}[/tex]
is continuous at x = 0.
[tex]\\\\\huge{⚝}[/tex] [tex]\\\\\huge{\underline{\mathcal{To Find:}}}[/tex]
[tex]\huge{→}[/tex]Value of k
[tex]\\\\\huge{⚝}[/tex] [tex]\\\\\huge{\underline{\mathcal{Steps}}}[/tex]
Here, f(0) = [tex]k + \dfrac{1}{2}[/tex]
[tex]\\\\[/tex]
First we will calculate for [tex]0^{-}[/tex]:
[tex]\\\large{⇒}[/tex][tex]\\\lim_{x\to {0}^{ - } }f(x) = \lim_{x\to {0}^{ - } } \: \dfrac{\sin \: 5x}{ {x}^{2} + 2x }\\[/tex]
[tex]\\\large{⇒}[/tex][tex]\\ \lim_{x\to {0}^{ - } }f(x) =\lim_{x\to {0}^{ - } } \dfrac{5 \times \sin5x}{5 \times x(x + 2)} \\ [/tex]
[tex]\\\large{⇒}[/tex][tex]\\\lim_{x\to {0}^{ - } }f(x) = \lim_{x\to {0}^{ - } } \dfrac{5}{x + 2} \times \lim_{x\to {0}^{ - } } \dfrac{ \sin5x }{5x} \\ [/tex]
[tex]\\\large{⇒}[/tex][tex] \\\lim_{x\to {0}^{ - } }f(x) = \lim_{x\to {0}^{ - } } \dfrac{5}{x + 2} \times 1\\[/tex]
[tex]\\\large{⇒}[/tex][tex]\\\lim_{x\to {0}^{ - } }f(x) =\lim_{x\to {0}^{ - } } \dfrac{5}{x + 2} \\[/tex]
[tex]\\\large{⇒}[/tex][tex]\\\lim_{x\to {0}^{ - } }f(x) = \dfrac{5}{0 + 2}\\[/tex]
[tex]\\\large{⇒}[/tex][tex] \\\lim_{x\to {0}^{ -} }f(x) = \dfrac{5}{2}\\\\ [/tex]
Now we will calculate for [tex]0^{+}[/tex]
[tex]\\\large{⇒}[/tex][tex]\\\lim_{x\to {0}^{ + } }f(x) = \lim_{x\to {0}^{ + } } \: \dfrac{\sin \: 5x}{ {x}^{2} + 2x }\\[/tex]
[tex]\\\large{⇒}[/tex][tex]\\\lim_{x\to {0}^{ + } }f(x) =\lim_{x\to {0}^{ + } } \dfrac{5 \times \sin5x}{5 \times x(x + 2)}\\ [/tex]
[tex]\\\large{⇒}[/tex][tex]\\\lim_{x\to {0}^{ + } }f(x) = \lim_{x\to {0}^{ + } } \dfrac{5}{x + 2} \times \lim_{x\to {0}^{ + } } \dfrac{ \sin5x }{5x}\\ [/tex]
[tex]\\\large{⇒}[/tex][tex]\\ \lim_{x\to {0}^{ + } }f(x) = \lim_{x\to {0}^{ + } } \dfrac{5}{x + 2} \times 1\\[/tex]
[tex]\\\large{⇒}[/tex][tex]\\ \lim_{x\to {0}^{ + } }f(x) = \lim_{x \to {0}^{ + } } \dfrac{5}{x + 2}\\[/tex]
[tex]\\\large{⇒}[/tex][tex] \\\lim_{x\to {0}^{ + } }f(x) = \dfrac{5}{0 + 2}\\[/tex]
[tex]\\\large{⇒}[/tex][tex]\\\lim_{x\to {0}^{ + } }f(x) = \dfrac{5}{2}\\\\[/tex]
For f(x) to be continuous at x = 0,
[tex]\large{⇒}[/tex][tex]\lim_{x\to {0}^{ - } } = \lim_{x\to {0}^{ + } } = \lim_{x\to 0}\\ [/tex]
[tex]\\\large{⇒}[/tex][tex] \\\lim_{x\to {0}^{ - } } = \lim_{x\to {0}^{ + } } = \dfrac{5}{ 2}\\ [/tex]
[tex]\\\large{⇒}[/tex][tex]\\ \lim_{x\to {0}^{ - } } = \lim_{x\to{0}}\\[/tex]
[tex]\\\large{⇒}[/tex][tex]\\ \lim_{x\to {0}^{ - } } = k + \dfrac{1}{2}\\ [/tex]
[tex]\\\large{⇒}[/tex][tex]\\ \dfrac{5}{2} = k + \dfrac{1}{2}\\ [/tex]
[tex]\\\large{⇒}[/tex][tex]\\ \dfrac{5}{2} - \dfrac{1}{2} = k \\[/tex]
[tex]\\\large{⇒}[/tex][tex]\\ k = \dfrac{4}{2}\\[/tex]
[tex]\large{⇒}[/tex][tex] k = 2[/tex]
[tex]\\\\\fbox{\fbox{\fbox{\fbox{\huge{\underline{\underline{\sf{\green{Hope\:it\:helps}}}}}}}}}[/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex]\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}[/tex]
is continuous at x = 0.
Consider,
[tex]\rm :\longmapsto\:f(0) = k + \dfrac{1}{2} [/tex]
Now,
Consider,
[tex]\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \tt f(x)}[/tex]
[tex]\rm \: = \: \: \displaystyle\lim_{x \to 0} \tt \dfrac{sin5x}{ {x}^{2} + 2x} [/tex]
If we substitute directly x = 0, we get indeterminant form.
So,
can be further rewritten as
[tex]\rm \: = \: \: \displaystyle\lim_{x \to 0} \tt \dfrac{sin5x}{ x \: ({x} + 2)} [/tex]
can be rewritten as
[tex]\rm \: = \: \: \displaystyle\lim_{x \to 0} \tt \dfrac{1}{x + 2} \times \displaystyle\lim_{x \to 0} \tt \dfrac{sin5x}{x} [/tex]
[tex]\rm \: = \: \: \tt \dfrac{1}{0+ 2} \times \displaystyle\lim_{x \to 0} \tt \dfrac{sin5x}{5x} \times 5[/tex]
We know,
[tex]\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{sinx}{x} = 1}}[/tex]
So, using this, we get
[tex]\rm \: = \: \: \dfrac{1}{2} \times 1 \times 5[/tex]
[tex]\rm \: = \: \: \dfrac{5}{2} [/tex]
So,
[tex]\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \tt f(x) \: = \: \dfrac{5}{2} }[/tex]
Now,
We know, A function f(x) is said to be continuous at x = a, iff
[tex]\red{\rm :\longmapsto\:\displaystyle\lim_{x \to a} \tt f(x) = f(a)}[/tex]
Now, According to statement, it is given that function f(x) is continuous at x = 0.
So,
[tex]\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \tt f(x) = f(0)}[/tex]
[tex]\rm :\longmapsto\:\dfrac{5}{2} = k + \dfrac{1}{2} [/tex]
[tex]\rm :\longmapsto\:k \: = \: \dfrac{5}{2} - \dfrac{1}{2} [/tex]
[tex]\rm :\longmapsto\:k \: = \: \dfrac{5 - 1}{2} [/tex]
[tex]\rm :\longmapsto\:k \: = \: \dfrac{4}{2} [/tex]
[tex]\bf\implies \:k \: = \: 2[/tex]
Additional Information :-
[tex]\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{sinx}{x} = 1}}[/tex]
[tex]\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{tanx}{x} = 1}}[/tex]
[tex]\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{log(1 + x)}{x} = 1}}[/tex]
[tex]\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{ {e}^{x} - 1}{x} = 1}}[/tex]
[tex]\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{ {a}^{x} - 1}{x} = loga}}[/tex]
[tex]\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{sin^{ - 1} x}{x} = 1}}[/tex]
[tex]\boxed{ \rm{ \displaystyle\lim_{x \to 0} \tt \frac{tan^{ - 1} x}{x} = 1}}[/tex]