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This is the second of our many quizzes on Determining limits using substitution. Click on this to access more such competitive quizzes on Calculus.

Determining limits using substitution Quiz 2

mathematics

 Determine the value of  $\lim_{x \to 0}\left( {{e}^{x}}+x{{e}^{2}} \right)+\underset{x\to 1}{\mathop{\lim }}\,\left( {{x}^{2}}+3 \right)$.

 Simplify the terms and then separately apply limits to each term and then simplify the terms according to the limit.

  $L=\lim_{x \to 0}\left( {{e}^{x}}+x{{e}^{2}} \right)+\underset{x\to 1}{\mathop{\lim }}\,\left( {{x}^{2}}+3 \right)\\ L=\lim_{x \to 0}{{e}^{x}}+\underset{x\to 0}{\mathop{\lim x{{e}^{2}}}}\,+\underset{x\to 1}{\mathop{\lim }}\,{{x}^{2}}+\underset{x\to 1}{\mathop{\lim }}\,3\\ \Rightarrow 1+0+1+3\\ \Rightarrow 5$ 

 What is the value of twice of $\underset{x\to 2}{\mathop{\lim }}\,\left| x-2 \right|x-2$.

 First simplify the function so as to obtain independent expression and then apply limits to the function and solve.

  $ L=\underset{x\to 2}{\mathop{\lim }}\,\left| x-2 \right|x-2 \\ \text{As }x \to 2\\ |x-2|\to 0 \\ \text{So, } x|x-2|\to 0\\ L=-2\\ L\times 2=-4$ 

 Find $\lim_{x \to 3}\left( \dfrac{{{x}^{2}}-x+\dfrac{1}{4}}{\left( x-\dfrac{1}{2} \right)\left( x+\dfrac{1}{2} \right)} \right)$.

 First simplify the expression such that it is converted into simpler form and then put the value of x as 3 and then solve to get the limit.

  $ L=\lim_{x \to 3}\left( \dfrac{{{x}^{2}}-x+\dfrac{1}{4}}{\left( x-\dfrac{1}{2} \right)\left( x+\dfrac{1}{2} \right)} \right)\\ L=\lim_{x \to 3}\left( \dfrac{{{\left( x-\dfrac{1}{2} \right)}^{2}}}{\left( x-\dfrac{1}{2} \right)\left( x+\dfrac{1}{2} \right)} \right)\\ L=\lim_{x \to 3}\left( \dfrac{\left( x-\dfrac{1}{2} \right)}{\left( x+\dfrac{1}{2} \right)} \right)\\ \Rightarrow \left( \dfrac{\left( 3-\dfrac{1}{2} \right)}{\left( 3+\dfrac{1}{2} \right)} \right)\\ \Rightarrow \dfrac{5}{7}$ 

 What is the  ${{\dfrac{1}{4}}^{th}}$ of the limit of  $\dfrac{\left( x+2 \right)\left( x+3 \right)}{\left( x+4 \right)\left( x+1 \right)}$  when $x\to \infty $.

 First simplify the term and then use the limit$\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{x}=0$ put this limit as zero and then further simplify to get the limit.

  $ L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( x+2 \right)\left( x+3 \right)}{\left( x+4 \right)\left( x+1 \right)}\\ =\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( 1+\dfrac{2}{x} \right)\left( 1+\dfrac{3}{x} \right)}{\left( 1+\dfrac{4}{x} \right)\left( 1+\dfrac{1}{x} \right)}=\dfrac{\left( 1+0 \right)\left( 1+0 \right)}{\left( 1+0 \right)\left( 1+0 \right)}=1\\ \Rightarrow \dfrac{1}{4}\times 1=\dfrac{1}{4}$ 

 First find the limits of every part using the properties of limit and then compare the value of the limits obtained to see in which one it is closest to zero.

  $ {{L}_{1}}=\lim_{x \to 0}\left( \dfrac{{{x}^{2}}}{1-\cos x} \right)\\ \Rightarrow \lim_{x \to 0}2\left( \dfrac{{{\left( \dfrac{x}{2} \right)}^{2}}}{\sin {{\left( \dfrac{x}{2} \right)}^{2}}} \right)\\ \Rightarrow 2\\ {{L}_{2}}=\underset{x\to 1}{\mathop{\lim }}\,\left( \dfrac{3x}{x-1}-\dfrac{3}{\ln (x)} \right) \\ \text{Apply L'Hospital's Rule: } \\  \text{=}\underset{x\to 1}{\mathop{\lim }}\,\left( \dfrac{3}{1}-\dfrac{3}{x} \right)\\ \Rightarrow 3-3=0\\ {{L}_{3}}=\underset{x\to 1}{\mathop{\lim }}\,\left( \dfrac{{{x}^{2}}+2x-8}{\sqrt{{{x}^{2}}+5}-\left( x+1 \right)} \right)\\ \Rightarrow \underset{x\to 1}{\mathop{\lim }}\,\left( \dfrac{1+2-8}{\sqrt{1+5}-\left( 1+1 \right)} \right)\\ \Rightarrow \dfrac{-6}{\sqrt{6}-2}\\ {{L}_{4}}=\underset{x\to \infty }{\mathop{\lim }}\,({{x}^{4}}+5{{x}^{3}}+x+4) \\ \text{As, }x\to \infty,  f\left( x \right)\to \infty \\ {{L}_{4}}=\infty $ 