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This is one of our many quizzes on Estimating limit values from tables. Click on this to access more such competitive quizzes on Limits.

Estimating limit values from tables Quiz 2

mathematics

 Estimate the value of $\underset{x\to 2}{\mathop{\lim }}\,\left( 3x+1 \right)$ using the given table $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/106cb83c-a784-4a25-bcf1-9aa4d05c701b8353459802385063897.png"/>

 Using the concept of estimating the limit from the table, try to find the value to which $f\left( x \right)$ tends, when $x$ approaches $2$.

 From the above table, we have to estimate the limit when $x$ tends to $2$. Given $f\left( x \right)=\left( 3x+1 \right)$. It is clear from the table that, when $x$ approaches $2$, $f\left( x \right)$ tends to $7$ approximately. Hence, the answer is $7$.

 Find the limit $\underset{x\to 0}{\mathop{\lim }}\,\left( \left| 2x \right| \right)$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/454efcd3-6e6c-4946-a6fc-42f5ce8157be5380875990127675118.png"/>

 Using the concept of estimating the limit from the table, try to find the value to which $f\left( x \right)$ tends, when $x$ approaches $0$.

  From the above table, we have to estimate the limit when $x$ tends to $0$. $ \\ $  Given $f\left( x \right)=\left| 2x \right|$. $ \\ $ It is clear from the table that, when $x$ approaches $0$, $f\left( x \right)$ tends to $0$ approximately. Hence, the answer is $0$.

 If the value of the limit $\underset{x\to \,-1}{\mathop{\lim }}\,\left( {{x}^{2}}+1 \right)$ is the first term of G.P which has a common ratio of $0.5$. Find the value of the sum of infinite terms of G.P. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9420a4c0-c9a5-4a0e-963a-8b9f987e3be67498775482088164889.png"/>

 First find the value of limit using the concept of estimating the limit from the table to find the value of the first term of the G.P. and then use the formula to find the sum of infinite terms of G.P.

Given $f\left( x \right)={{x}^{2}}+1$. It is clear from the table that, when $x$ approaches $-1$ , $f\left( x \right)$ tends to $2$ approximately. $ \\ $ Hence the value of the limit $\underset{x\to \,-1}{\mathop{\lim }}\,\left( {{x}^{2}}+1 \right)$  is $2$. $ \\ $ Therefore, $2$ is the first term of G.P with a common ratio $0.5$. $ \\ $ The sum of infinite terms of a G.P is  $\dfrac{a}{1-r}$ , where $a$ is the first term and $r$ is the common ratio of the G.P. $ \\ $ Here $a=2$ , $r=0.5$ $ \\ {{S}_{\infty }}=\dfrac{a}{1-r}=\dfrac{2}{1-0.5}$ $\therefore {{S}_{\infty }}=4$

 If the value of $\underset{x\to \,-1}{\mathop{\lim }}\,\left( 2{{x}^{2}}-4 \right)$ is $\alpha $. Then find one of the roots of the quadratic equation ${{x}^{2}}+\left( 3\alpha  \right)x-4\alpha =0$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/98881c8a-2766-4e69-a3c4-91322f8d99ba4769057770116282306.png"/>

 Find the value of the given limit to find the value of $\alpha $, then substitute it in the given equation to find the roots of it.

 Given $f\left( x \right)=2{{x}^{2}}-4$ $ \\ $ It is clear from the table that, when $x$ approaches $-1$ , $f\left( x \right)$ tends to $-2$ approximately. $ \\ $ Hence the value of the limit $\underset{x\to \,-1}{\mathop{\lim }}\,\left( 2{{x}^{2}}-4 \right)$  is $-2$. $ \\ \therefore \alpha =-2 \\ $ Given quadratic equation ${{x}^{2}}+\left( 3\alpha  \right)x-4\alpha =0$ $ \\ $ Substituting the value of $\alpha $ in the above equation , we get , $ \\ {{x}^{2}}-6x+8=0 \\ \Rightarrow {{x}^{2}}-4x-2x+8=0 \\ \Rightarrow \left( x-2 \right)\left( x-4 \right)=0 \\ \therefore x=2$ or $x=4$

 If the value of limit $\underset{x\to \,\,3}{\mathop{\lim }}\,\left( \dfrac{{{x}^{2}}-5x+6}{x-3} \right)$ is $a$. If $a={{\sin }^{2}}\theta +\dfrac{\cos \left( {{150}^{\circ }} \right)+1}{2}$ then find the possible value of $\theta $ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7f1b2ce5-f509-479c-9bb5-12bf7556c72b4501024043399406822.png"/>

 Find the value of $a$ by finding the limit value, then substitute it in the given equation and then simplify it to find the value of $\theta $.

 Given $f\left( x \right)=\dfrac{{{x}^{2}}-5x+6}{x-3}$ $ \\ $ It is clear from the table that, when $x$ approaches 3 , $f\left( x \right)$ tends to $1$  approximately. $ \\ $ Hence the value of the limit $\underset{x\to \,\,3}{\mathop{\lim }}\,\left( \dfrac{{{x}^{2}}-5x+6}{x-3} \right)$  is $1$. $ \\ \therefore a=1 \\ $ Given $a={{\sin }^{2}}\theta +\dfrac{\cos \left( {{150}^{\circ }} \right)+1}{2}$ $ \\ $ Substituting the value of $a$ , we get , $ \\ 1={{\sin }^{2}}\theta +\dfrac{\cos \left( {{150}^{\circ }} \right)+1}{2} \\ \Rightarrow \dfrac{2{{\sin }^{2}}\theta +\cos \left( {{150}^{\circ }} \right)+1}{2}=1 \\ \Rightarrow 2{{\sin }^{2}}\theta =1-\cos \left( {{150}^{\circ }} \right) \\ $ As we know that , $1-\cos \theta =2{{\sin }^{2}}\left( \dfrac{\theta }{2} \right)$ $ \\ \Rightarrow 2{{\sin }^{2}}\theta =2{{\sin }^{2}}{{\left( \dfrac{150}{2} \right)}^{\circ }} \\ \Rightarrow \theta ={{75}^{\circ }}$