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Does the limit exist? If yes, find its limit. If the limit does not exist, determine if it is...

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Does the limit exist? If yes, find its limit. If the limit does not exist, determine if it is infinity, minus infinity or neither

{eq}a. \ \lim_{t\rightarrow 0}\frac{1}{\sqrt[t]{1+5t}}-\frac{1}{t}\\ \\ \\ b.\ \lim_{x\rightarrow \infty }\frac{\sqrt{2x^4+5x^3+x^2-1}}{6x^2+1000}\\ \\ c.\ \lim_{x\rightarrow 0}\frac{x^7-\sin ^2x}{x^2}\\ \\ d. \ \lim_{x\rightarrow \infty }x^2-x {/eq}

Limit:


Suppose that F(t) be any function of t then limit of the function F(t) at {eq}\displaystyle t \to \alpha {/eq} is defined by

{eq}\displaystyle \mathop {\lim }\limits_{t \to \alpha } F(t) = F(\alpha ) = L. {/eq}

When L is finite then limit of the function exists and when L is plus infinity or minus infinity then limit does not exists.


Required formulas:

{eq}\displaystyle \eqalign{ & 1.\,\,\,\mathop {\lim }\limits_{x \to \infty } {a^x} = \infty \cr & 2.\,\,\mathop {\lim }\limits_{x \to \infty } x = \infty \cr & 3.\,\,\mathop {\lim }\limits_{x \to 0} \frac{1}{x} = \infty \cr & 4.\,\,\mathop {\lim }\limits_{x \to 0} \frac{1}{{{{(ax + b)}^{1/x}}}} = 0 \cr & 5.\,\,\cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - .... \cr & 6.\,\,\cos 2x = 1 - 2{\sin ^2}x \cr} {/eq}


Answer and Explanation:


(a).

{eq}\displaystyle \eqalign{ & \mathop {\lim }\limits_{t \to 0} \left( {\frac{1}{{{{\left( {1 + 5t} \right)}^{1/t}}}} - \frac{1}{t}} \right) = \mathop {\lim }\limits_{t \to 0} \frac{1}{{{{\left( {1 + 5t} \right)}^{1/t}}}} - \mathop {\lim }\limits_{t \to 0} \frac{1}{t} \cr & \mathop {\lim }\limits_{t \to 0} \left( {\frac{1}{{\sqrt {1 + 5t} }} - \frac{1}{t}} \right) = \frac{1}{{{{\left( {1 + 5t} \right)}^\infty }}} - \infty \cr & \mathop {\lim }\limits_{t \to 0} \left( {\frac{1}{{\sqrt {1 + 5t} }} - \frac{1}{t}} \right) = \frac{1}{\infty } - \infty \cr & \mathop {\lim }\limits_{t \to 0} \left( {\frac{1}{{\sqrt {1 + 5t} }} - \frac{1}{t}} \right) = 0 - \infty \cr & \mathop {\lim }\limits_{t \to 0} \left( {\frac{1}{{\sqrt {1 + 5t} }} - \frac{1}{t}} \right) = - \infty \cr} {/eq}

Hence the limit does not exists.


(b).

{eq}\displaystyle \eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {2{x^4} + 5{x^3} + {x^2} - 1} }}{{6{x^2} + 1000}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\sqrt {2 + \frac{5}{x} + \frac{1}{{{x^2}}} - \frac{1}{{{x^4}}}} }}{{{x^2}\left( {6 + \frac{{1000}}{{{x^2}}}} \right)}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {2{x^4} + 5{x^3} + {x^2} - 1} }}{{6{x^2} + 1000}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {2 + \frac{5}{x} + \frac{1}{{{x^2}}} - \frac{1}{{{x^4}}}} }}{{\left( {6 + \frac{{1000}}{{{x^2}}}} \right)}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {2{x^4} + 5{x^3} + {x^2} - 1} }}{{6{x^2} + 1000}} = \frac{{\sqrt {2 + 0 + 0 - 0} }}{{\left( {6 + 0} \right)}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {2{x^4} + 5{x^3} + {x^2} - 1} }}{{6{x^2} + 1000}} = \frac{{\sqrt 2 }}{6}. \cr} {/eq}


(c).

{eq}\displaystyle \eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7}}}{{{x^2}}} - \mathop {\lim }\limits_{x \to \infty } \frac{{{{\sin }^2}x}}{{{x^2}}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7}}}{{{x^2}}} - \mathop {\lim }\limits_{x \to \infty } \frac{{{{\sin }^2}x}}{{{x^2}}} \cr & \left[ {{{\sin }^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)} \right] \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \infty } {x^5} - \frac{1}{2}\mathop {\lim }\limits_{x \to \infty } \frac{{\left( {1 - \cos 2x} \right)}}{{{x^2}}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \infty } {x^5} - \frac{1}{2}\mathop {\lim }\limits_{x \to \infty } \frac{{\left( {1 - \left( {1 - \frac{{{{(2x)}^2}}}{{2!}} + \frac{{{{(2x)}^4}}}{{4!}} - .....} \right)} \right)}}{{{x^2}}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \infty } {x^5} - \frac{1}{2}\mathop {\lim }\limits_{x \to \infty } \frac{{\left( {2{x^2} - \frac{{2{x^4}}}{3} + .....} \right)}}{{{x^2}}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \infty } {x^5} - \frac{1}{2}\mathop {\lim }\limits_{x \to \infty } \left( {2 - \frac{{2{x^2}}}{3} + ....} \right) \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \infty } {x^5} - \left( {1 + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{2{x^2}}}{3} - ..........} \right)} \right) \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \infty - 1 + \infty \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^7} - {{\sin }^2}x}}{{{x^2}}} = \infty . \cr} {/eq}

Hence the limit does not exists.


(d). {eq}\displaystyle \eqalign{ & \mathop {\lim }\limits_{x \to \infty } \left( {{x^2} - x} \right) = \mathop {\lim }\limits_{x \to \infty } x\left( {x - 1} \right) \cr & \mathop {\lim }\limits_{x \to \infty } \left( {{x^2} - x} \right) = \infty .\cr} {/eq}

Hence the limit does not exists.



Learn more about this topic:

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How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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