Question

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$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {\lim_{h \rightarrow 0}\: \dfrac{1}{h} \bigg\lgroup \dfrac{1}{\sqrt{x + h}} - \dfrac{1}{\sqrt{x}} \bigg\rgroup }$
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Answer 1

Answer:

Answer = -1 /2x√x

Step-by-step explanation:

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\rm \: {\lim_{h \rightarrow 0}\: \dfrac{1}{h} \bigg\lgroup \dfrac{1}{\sqrt{x + h}} - \dfrac{1}{\sqrt{x}} \bigg\rgroup }$

On substituting the value of x = 0, we get indeterminant form, so we can't solve it directly by direct substitution.

Take out LCM, we get

$= \rm \: {\lim_{h \rightarrow 0}\: \dfrac{1}{h} \bigg\lgroup \dfrac{ \sqrt{x} - \sqrt{x + h} }{\sqrt{x + h} \sqrt{x} } \bigg\rgroup }$

$= \rm \lim_{h \rightarrow 0}\dfrac{1}{ \sqrt{x} \sqrt{x + h} } \times \rm \: \lim_{h \rightarrow 0}\dfrac{ \sqrt{x} - \sqrt{x + h} }{h}$

$= \rm \: \dfrac{1}{ \sqrt{x} } \times \dfrac{1}{ \sqrt{x} } \lim_{h \rightarrow 0}\dfrac{ \sqrt{x} - \sqrt{x + h} }{h} \times \dfrac{ \sqrt{x} + \sqrt{x + h} }{ \sqrt{x} + \sqrt{x + h} }$

$= \rm \: \dfrac{1}{x} \lim_{h \rightarrow 0}\dfrac{x - (x + h)}{h( \sqrt{x + h} + \sqrt{x}) }$

$= \rm \: \dfrac{1}{x} \lim_{h \rightarrow 0}\dfrac{ - \cancel h}{ \cancel{h}( \sqrt{x + h} + \sqrt{x}) }$

$= \rm \: \dfrac{1}{x} \lim_{h \rightarrow 0}\dfrac{ - 1}{( \sqrt{x + h} + \sqrt{x}) }$

$= \rm \: - \dfrac{1}{x} \times \dfrac{1}{ \sqrt{x} + \sqrt{x} }$

$= \rm \: - \dfrac{1}{2x \sqrt{x} }$

$\rm :\implies\: \boxed{ \pink{ \sf \:  \rm \: {\lim_{h \rightarrow 0}\: \dfrac{1}{h} \bigg\lgroup \dfrac{1}{\sqrt{x + h}} - \dfrac{1}{\sqrt{x}} \bigg\rgroup }\:  =  \tt \: - \dfrac{1}{2x \sqrt{x} } }}$

Additional Information :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{sin \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{tan \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ log(1 + x) }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {e}^{x} \: - \: 1}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {a}^{x} \: - \: 1}{x} \: = \: log(a) }}}}}} \\ \end{gathered}$