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

Introduction to limits Quiz 2

mathematics

 Evaluate $\mathop {\lim }\limits_{x \to 5} \dfrac{{{x^2} - 3x - 10}}{{2{x^2} - 9x - 5}}$

 Factorize the numerator and denominator to simplify and then evaluate the limit. 

$ &nbsp;&nbsp;\mathop {\lim }\limits_{x \to 5} \dfrac{{{x^2} - 3x - 10}}{{2{x^2} - 9x - 5}} = \mathop {\lim }\limits_{x \to 5} \dfrac{{{x^2} - 5x + 2x - 10}}{{2{x^2} + x - 10x - 5}} \\ &nbsp;&nbsp;&nbsp;= \mathop {\lim }\limits_{x \to 5} \dfrac{{x\left( {x - 5} \right) + 2\left( {x - 5} \right)}}{{x\left( {2x + 1} \right) - 5\left( {2x + 1} \right)}} = \mathop {\lim }\limits_{x \to 5} \dfrac{{\left( {x + 2} \right)\left( {x - 5} \right)}}{{\left( {2x + 1} \right)\left( {x - 5} \right)}} \\ &nbsp;&nbsp;&nbsp;= \mathop {\lim }\limits_{x \to 5} \dfrac{{\left( {x + 2} \right)}}{{\left( {2x + 1} \right)}} = \dfrac{7}{{11}} \\ &nbsp;$

 $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}x}}{{{x^2}}}$ is equal to

 Use L’Hospital’s Rule to calculate the limit.

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}x}}{{{x^2}}} = \dfrac{0}{0}$. So we can use the L’Hospital’s Rule. $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x\cos x}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{2x}}$ We know $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ Therefore, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{2x}} = 1$

 If $\dfrac{{\cos \left( {2x - 2} \right) + 1}}{2} \leqslant f\left( x \right) \leqslant 3{x^2} - 6x + 4$ for all real values of $x$, then $\mathop {\lim }\limits_{x \to 1} f\left( x \right)$ is equal to

 Use Sandwich Theorem to solve the problem.

$\mathop {\lim }\limits_{x \to 1} \dfrac{{\cos \left( {2x - 2} \right) + 1}}{2} = \dfrac{{\cos 0 + 1}}{2} = 1$ And, $\mathop {\lim }\limits_{x \to 1} 3{x^2} - 6x + 4 = 1$ From Sandwich Theorem, If $g\left( x \right) \leqslant f\left( x \right) \leqslant h\left( x \right)$ and $g\left( a \right) = h\left( a \right)$, then $f\left( a \right) = g\left( a \right) = h\left( a \right)$. Thus, $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 1$

 The value of $\mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin \left( {\dfrac{\pi }{{2x}}} \right)} \right)$ is

We know that, $ - 1 \leqslant \sin \left( {\dfrac{\pi }{{2\left( {x - 1} \right)}}} \right) \leqslant 1\;\;\forall x \in R$ Thus, $ - {x^2} \leqslant {x^2}\sin \left( {\dfrac{\pi }{{2\left( {x - 1} \right)}}} \right) \leqslant {x^2}$ $\mathop {\lim }\limits_{x \to 0} \left( {{x^2}} \right) = 0$ and $\mathop {\lim }\limits_{x \to 0} \left( { - {x^2}} \right) = 0$ Thus by Sandwich Theorem, $\mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin \left( {\dfrac{\pi }{{2x}}} \right)} \right) = 0$

 For a function $f\left( x \right)$, $\mathop {\lim }\limits_{x \to 3} f\left( x \right)$ exists. If given that $\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {16 - {{\left( {f\left( x \right)} \right)}^2}} }}{{\left| {3 - x} \right|}} = 2$, then the value of $\mathop {\lim }\limits_{x \to 3} f\left( x \right)$ is

 Check for the condition for applying L’Hospital’s Rule.

At $x = 3$, denominator of $\dfrac{{\sqrt {16 - {{\left( {f\left( x \right)} \right)}^2}} }}{{\left| {3 - x} \right|}}$ becomes $0$. To calculate $\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {16 - {{\left( {f\left( x \right)} \right)}^2}} }}{{\left| {3 - x} \right|}}$ we would have to use L’Hospital’s Rule. The condition for applying L’Hospital’s Rule is that $f\left( a \right)$ should be indeterminate. Since the numerator of $\dfrac{{\sqrt {16 - {{\left( {f\left( x \right)} \right)}^2}} }}{{\left| {3 - x} \right|}}$ cannot be $\infty $, it has to be zero at $x = 3$ to get indeterminate form. Thus, $ &nbsp;&nbsp;\sqrt {16 - {{\left( {f\left( 3 \right)} \right)}^2}} = 0 \\ &nbsp;&nbsp;&nbsp;\Rightarrow 16 - {\left( {f\left( 3 \right)} \right)^2} = 0 \\ &nbsp;&nbsp;&nbsp;\Rightarrow f\left( 3 \right) = 4 \\ &nbsp;$ Thus, $\mathop {\lim }\limits_{x \to 3} f\left( x \right) = f\left( 3 \right) = 4$