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Infinite limits Quiz 2

mathematics

 Evaluate $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{3+5x}-\sqrt{3-5x}}{3x}$.

 Rationalise $\dfrac{\sqrt{3+5x}-\sqrt{3-5x}}{3x}$ and then apply limits.

$\begin{align} &nbsp;&nbsp;& \text{Consider the given function} \\ &nbsp;& \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{3+5x}-\sqrt{3-5x}}{3x} \\ &nbsp;& \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{3+5x}-\sqrt{3-5x}}{3x}\times \dfrac{\sqrt{3+5x}+\sqrt{3-5x}}{\sqrt{3+5x}+\sqrt{3-5x}} \\ &nbsp;& \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{10}{3}.\dfrac{1}{\sqrt{3+5x}+\sqrt{3-5x}}=\dfrac{10}{3}.\dfrac{1}{2\sqrt{3}} \\ &nbsp;& \Rightarrow \dfrac{5\sqrt{3}}{9} \\ \end{align}$

 Evaluate  $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\cos 3x-\cos 7x}{{{x}^{2}}}$.

 Apply sum to product formula for  $\operatorname{cosA}-\operatorname{cosB}=2\dfrac{\sin A+B}{2}.\dfrac{\operatorname{sinB}-A}{2}$ and use $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$.

$\begin{align} &nbsp;&nbsp;& \text{Consider the given function} \\ &nbsp;& \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\cos 3x-\cos 7x}{{{x}^{2}}} \\ &nbsp;& \underset{x\to 0}{\mathop{\lim }}\,\dfrac{2\sin 5x.\sin 2x}{{{x}^{2}}} \\ &nbsp;& \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{2\sin 5x}{5x}.\dfrac{\sin 2x}{2x}\times 5\times 2 \\ &nbsp;& \Rightarrow 5\times 2\times 2\left[ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin 5x}{5x} \right]\left[ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin 2x}{2x} \right] \\ &nbsp;& \Rightarrow 20\times \left( 1 \right)\times \left( 1 \right)\left\{ \therefore \underset{\theta \to 0}{\mathop{\lim }}\,\dfrac{\sin \theta }{\theta }=1 \right\} \\ &nbsp;& \Rightarrow 20 \\ \end{align}$

 Find the value of $b$ if $\underset{x\to -b}{\mathop{\lim }}\,\dfrac{{{x}^{17}}+{{b}^{17}}}{x+b}=17$.

 Use formula $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$ to find the value of b.

$\begin{align} &nbsp;&nbsp;& \underset{x\to -b}{\mathop{\lim }}\,\dfrac{{{x}^{17}}+{{b}^{17}}}{x+b}=17 \\ &nbsp;& \Rightarrow \underset{x\to -b}{\mathop{\lim }}\,\dfrac{{{x}^{17}}-\left( -{{b}^{17}} \right)}{x+\left( -b \right)}=17 \\ &nbsp;& \Rightarrow 17{{\left( -b \right)}^{16}}=17 \\ &nbsp;& \Rightarrow b=\pm 1 \\ \end{align}$

 Solve $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1-\cos 9x}{1-\cos 7x}$

 Apply power reducing formula $1-{{\cos }^{2}}A=2{{\sin }^{2}}A$ and $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$.

$\begin{align} &nbsp;&nbsp;& \text{Given function} \\ &nbsp;& \underset{x\to 0}{\mathop{\lim }}\,\dfrac{1-\cos 9x}{1-\cos 7x} \\ &nbsp;& \underset{x\to 0}{\mathop{\lim }}\,\dfrac{2{{\sin }^{2}}\left( \dfrac{9x}{2} \right)}{2{{\sin }^{2}}\left( \dfrac{7x}{2} \right)}={{\left[ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \dfrac{9x}{2}}{\sin \dfrac{7x}{2}} \right]}^{2}} \\ &nbsp;& \Rightarrow {{\left[ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\sin \dfrac{9x}{2}}{\dfrac{9x}{2}}.\dfrac{9x}{2}}{\dfrac{\sin \dfrac{7x}{2}}{\dfrac{7x}{2}}.\dfrac{7x}{2}} \right]}^{2}} \\ &nbsp;& =\dfrac{{{9}^{2}}}{{{7}^{2}}} \\ &nbsp;& =\dfrac{81}{49} \\ \end{align}$

 Evaluate $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{7}^{x}}-{{5}^{x}}}{\sin x}$.

 Multiply and divide by $x$ and use $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{x}}-1}{x}=\log a$.

$\begin{align} &nbsp;&nbsp;& \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{7}^{x}}-{{5}^{x}}}{\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( {{7}^{x}}-1 \right)-\left( {{5}^{x}}-1 \right)}{x}\dfrac{x}{\sin x} \\ &nbsp;& \Rightarrow \left[ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{7}^{x}}-1}{x}-\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{5}^{x}}-1}{x} \right]\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x} \\ &nbsp;& \Rightarrow \left[ \log 7-\log 5 \right].1 \\ &nbsp;& =\log \dfrac{7}{5} \\ \end{align}$