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Properties of limits Quiz 2

mathematics

 Simplify the limit, $\underset{k\to -5}{\mathop{\lim }}\,\dfrac{{{23}^{k+5}}-1}{k+5}.$

 Apply $k+5=t$ and use a suitable limit formula to solve further.

$\begin{align} &nbsp;&nbsp;& \text{Consider the given limit }\underset{k\to -5}{\mathop{\lim }}\,\dfrac{{{23}^{k+5}}-1}{k+5}\text{.} \\ &nbsp;& \text{Put }k+5=t \\ &nbsp;& \Rightarrow k=t-5 \\ &nbsp;& \Rightarrow \underset{t-5\to -5}{\mathop{\lim }}\,\dfrac{{{23}^{t}}-1}{t} \\ &nbsp;& \Rightarrow \underset{t\to 0}{\mathop{\lim }}\,\dfrac{{{23}^{t}}-1}{t} \\ &nbsp;& \Rightarrow \log 23\left[ \because \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{x}}-1}{x}=\log a \right] \\ \end{align}$

 Let $x$ be the sum of first $'n'$ natural numbers and $y$ be the sum of squares of first $'n'$natural numbers. Write the value of $\underset{n\to 10}{\mathop{\lim }}\,x+\underset{n\to 1}{\mathop{\lim }}\,y.$

 Use formula for sum of first $n$ natural numbers and sum of squares of first $n$ natural numbers and find their limits.

$\begin{align} &nbsp;&nbsp;& \text{Given,} \\ &nbsp;& x=\text{ sum of first }n\text{ natural number} \\ &nbsp;& \Rightarrow x=\dfrac{n\left( n+1 \right)}{2} \\ &nbsp;& \text{Now, }\underset{x\to \infty }{\mathop{\lim }}\,x=\underset{n\to 10}{\mathop{\lim }}\,\dfrac{n\left( n+1 \right)}{2} \\ &nbsp;& \Rightarrow \dfrac{10\times 11}{2}=55 \\ &nbsp;& \text{Also given}, \\ &nbsp;& y=\text{ sum of squares of }n\text{ natural numbers} \\ &nbsp;& \Rightarrow y=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \\ &nbsp;& \Rightarrow \underset{n\to 1}{\mathop{\lim }}\,=\underset{n\to 1}{\mathop{\lim }}\,\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \\ &nbsp;& \Rightarrow \dfrac{1\times 2\times 3}{6}=1 \\ &nbsp;& \therefore \underset{n\to 10}{\mathop{\lim }}\,x+\underset{n\to 1}{\mathop{\lim }}\,y=55+1=56 \\ \end{align}$

 Let $x=\underset{\theta \to \dfrac{\pi }{2}}{\mathop{\lim }}\,{{\sin }^{-1}}\left( \dfrac{2\tan \dfrac{\theta }{4}}{1+{{\tan }^{2}}\dfrac{\theta }{4}} \right),$ then evaluate $\underset{y\to x}{\mathop{\lim }}\,\cos y.$

 First solve $x$ by finding the limit where $\theta \to \dfrac{\pi }{2}$. Then put this value of $x$ in the required limit to be obtained. 

$\begin{align} &nbsp;&nbsp;& \text{Consider, }x=\underset{\theta \to \dfrac{\pi }{2}}{\mathop{\lim }}\,{{\sin }^{-1}}\left( \dfrac{2\tan \dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{4}} \right) \\ &nbsp;& \Rightarrow x=\underset{\theta \to \dfrac{\pi }{2}}{\mathop{\lim }}\,{{\sin }^{-1}}\sin \left( 2\times \dfrac{\theta }{4} \right)\left[ \because \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta \right] \\ &nbsp;& \Rightarrow x=\underset{\theta \to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{\theta }{2}=\dfrac{\pi }{4} \\ &nbsp;& \text{Now, }\underset{y\to x}{\mathop{\lim }}\,\cos y=\underset{y\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\cos y \\ &nbsp;& \Rightarrow \cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} \\ \end{align}$

 Evaluate: $\underset{b\to 0}{\mathop{\lim }}\,\dfrac{{{\left( \dfrac{3}{2} \right)}^{b}}-{{\left( \dfrac{5}{7} \right)}^{b}}+\sin \left( \dfrac{3b}{2} \right)}{b}.$

 First add and subtract $1$ in the numerator and then use a suitable limit to solve further. 

$\begin{align} &nbsp;&nbsp;& \text{Consider the given limit, }\underset{b\to 0}{\mathop{\lim }}\,\dfrac{{{\left( \dfrac{3}{2} \right)}^{b}}-{{\left( \dfrac{5}{7} \right)}^{b}}+\sin \left( \dfrac{3b}{2} \right)}{b} \\ &nbsp;& \text{Add and subtract }1\text{ in the numerator} \\ &nbsp;& \Rightarrow \underset{b\to 0}{\mathop{\lim }}\,\dfrac{\left[ {{\left( \dfrac{3}{2} \right)}^{b}}-1 \right]-\left[ {{\left( \dfrac{5}{7} \right)}^{b}}-1 \right]+\sin \left( \dfrac{3b}{2} \right)}{b} \\ &nbsp;& \Rightarrow \underset{b\to 0}{\mathop{\lim }}\,\dfrac{{{\left( \dfrac{3}{2} \right)}^{b}}-1}{b}-\underset{b\to 0}{\mathop{\lim }}\,\dfrac{{{\left( \dfrac{5}{7} \right)}^{b}}-1}{b}+\underset{b\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{3b}{2} \right)}{\dfrac{3b}{2}}\times \dfrac{3}{2} \\ &nbsp;& \text{Use }\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{x}}-1}{x}=\log a\text{ and }\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 \\ &nbsp;& \Rightarrow \log \left( \dfrac{3}{2} \right)-\log \left( \dfrac{5}{7} \right)+\dfrac{3}{2} \\ &nbsp;& \Rightarrow \dfrac{3}{2}+\log \left( \dfrac{21}{10} \right) \\ \end{align}$

 Write the value of the following limits $\underset{y\to 0}{\mathop{\lim }}\,\dfrac{\sin \left[ y.\left( {{\sin }^{-1}}y+{{\cos }^{-1}}y \right) \right]}{y}+\underset{g\to 0}{\mathop{\lim }}\,\dfrac{\tan \left( \dfrac{29g}{3} \right)}{g}.$

 First use ${{\sin }^{-1}}\theta +{{\cos }^{-1}}\theta =\dfrac{\pi }{2}$ and reduce the given limit in simplest form. Then by using suitable limit formulae, solve further. 

$\begin{align} &nbsp;&nbsp;& \text{Given,} \\ &nbsp;& \underset{y\to 0}{\mathop{\lim }}\,\dfrac{\sin \left[ y.\left( {{\sin }^{-1}}y+{{\cos }^{-1}}y \right) \right]}{y}+\underset{g\to 0}{\mathop{\lim }}\,\dfrac{\tan \left( \dfrac{29}{3}g \right)}{g} \\ &nbsp;& \text{The above can be written as,} \\ &nbsp;& \Rightarrow \underset{y\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{\pi }{2}y \right)}{y}+\underset{g\to 0}{\mathop{\lim }}\,\dfrac{\tan \left( \dfrac{29}{3}g \right)}{g} \\ &nbsp;& \text{Solve further} \\ &nbsp;& \Rightarrow \underset{y\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{\pi y}{2} \right)}{\dfrac{\pi y}{2}}\times \dfrac{\pi }{2}+\underset{g\to 0}{\mathop{\lim }}\,\dfrac{\tan \left( \dfrac{29g}{3} \right)}{\dfrac{29g}{3}}\times \dfrac{29}{3} \\ &nbsp;& \Rightarrow \dfrac{\pi }{2}+\dfrac{29}{3}\left[ \because \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1,\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan x}{x}=1 \right] \\ &nbsp;& \Rightarrow \dfrac{3\pi +58}{6} \\ \end{align}$