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Limit Quiz 2

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 Determine   $\underset{x\to 0 }{\mathop{\lim }}\,\left( \dfrac{\left( x+3 \right)\left( x+1 \right)}{\left( x+2 \right)\left( x+4 \right)} \right)$

 Divide the whole function by $\dfrac{1}{x}$ and then apply $\underset{x\to 0}{\mathop{\lim }}\,$ therefore we see that all the terms having x as denominator becomes zero substitute this and obtain the solution.

  $\underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{(x+3)(x+1)}{(x+2)(x+4)} \right) \\ \text{Divide by }\dfrac{1}{x}\text{ to both denominator and numerator} \\ \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{x(1+\dfrac{3}{x})(1+\dfrac{1}{x})}{x(1+\dfrac{2}{x})(1+\dfrac{4}{x})} \right) \\ \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{(1+\dfrac{3}{x})(1+\dfrac{1}{x})}{(1+\dfrac{2}{x})(1+\dfrac{4}{x})} \right) \\ \text{Since for }\dfrac{3}{x},\dfrac{1}{x},\dfrac{2}{x},\dfrac{4}{x}\underset{x\to 0}{\mathop{\lim }}\,\text{ will be zero }\text{.} \\ \text{therefore,} \\ \Rightarrow \dfrac{(1+0)}{1+0}=1$ 

 What is the nature of $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{x}$

 First use the limit property and separate the limits between the two functions and then apply the limit to the individual function to get the value.

  $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{x} \\ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{\sqrt{1+x}}+\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{x} \\ \text{As x}\to \text{0  }\dfrac{1}{\sqrt{x+1}}\to 1\text{ but }\dfrac{1}{x}\to \infty  \\ \text{So, diverging nature }$

 What is the limit of  $\dfrac{\sin \left( x-2 \right)}{{{x}^{2}}+2x-8}$ at $x=2$.

 Apply the L Hospital’s Rule which means differentiating both the numerator and denominator and therefore after applying put the limits .

  $\text{Applying L'Hospital }\!\!'\!\!\text{ s Rule ( Note: }\dfrac{0}{0}or\dfrac{\infty }{\infty }\text{ form)} \\ \Rightarrow \dfrac{\sin \left( x-2 \right)}{{{x}^{2}}+2x-8}(\text{Differentiate both denominator and numerator)} \\ \underset{x\to 2}{\mathop{\lim }}\,\dfrac{\cos (x-2)}{2x+2}=\dfrac{\cos (2-2)}{2\times 2+2}=\dfrac{1}{6}$

 What is thrice the limit of $f(x)=\sqrt{4{{x}^{2}}+x}-2x$ at  $x=\infty $.

 First multiply by $(\sqrt{4{{x}^{2}}+x}+2x)$ to both numerator and denominator and then simply the equation and then solve it using limits.

  $\text{Let, }L= \sqrt{4{{x}^{2}}+x}-2x \\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{(\sqrt{4{{x}^{2}}+x}-2x)(\sqrt{4{{x}^{2}}+x}+2x)}{\sqrt{4{{x}^{2}}+x}+2x} \\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{(\sqrt{4{{x}^{2}}+x})}^{2}}-4{{x}^{2}}}{\sqrt{4{{x}^{2}}+x}+2x} \\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{\dfrac{\sqrt{4{{x}^{2}}+x}+2x}{x}} \\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{\sqrt{4+\dfrac{1}{x}}+2} \\ \text{As, }\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{x}=0 \\ L=\dfrac{1}{4} \\ \text{Thrice the value of L is }\dfrac{3}{4}$

 What is the difference between the limits of functions  $f(x)=\sqrt{\dfrac{8{{x}^{2}}-3}{2{{x}^{2}}+x}}$ and  $f(x)=\dfrac{2\cos x+\sin x}{x}$ at  $x=\infty $.

 Evaluate both the limits by using the limit property and after that obtain the difference between the two limits.

  $\text{As }f(x)=\sqrt{\dfrac{8{{x}^{2}}-3}{2{{x}^{2}}+x}} \\ {{L}_{1}}=\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{\dfrac{8-\dfrac{3}{{{x}^{2}}}}{2+\dfrac{1}{x}}} \\ \text{Now since }\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{x}=0 \\ {{L}_{1}}=\sqrt{4}=2 \\ \text{Now } \\ {{L}_{2}}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{2\cos x+\sin x}{x} \\ \text{From the squeeze theorem we know that }\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\cos x}{x}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\sin x}{x}=0 \\ {{L}_{2}}=0 \\ \text{The difference between limits =}{{\text{L}}_{1}}-{{L}_{2}}=2-0 =2$