Find the limits lim \ t\rightarrow 0 [( \frac {sin(5t)}{sin(4t)}) i+( \frac {ln(sin(5t))}{...

Question:

Find the limits

{eq}lim \ t\rightarrow 0 [( \frac {sin(5t)}{sin(4t)}) i+( \frac {ln(sin(5t))}{ ln(tan(4t))}) j+t \ ln(5t)k]= \\ lim \ t\rightarrow 5 [ (\frac {t^3 -125}{ t-5)} i+( \frac {t^2 -9t+20}{ t^2 -3t-10}) j+(t-5)ln(t-5)k= {/eq}

Existence of a Limit:

Suppose that {eq}a(x) {/eq} is a function that is defined in an interval that contains {eq}x=t {/eq}. Then limit is defined as:

{eq}\mathop {\lim }\limits_{x \to t} a(x) = M {/eq}

There exists a very small number {eq}k {/eq} such that {eq}k>0 {/eq} so that {eq}n>0 {/eq}. This means that {eq}|a(x) - M| < k {/eq} whenever {eq}0 < |x - t| < n {/eq}.

if {eq}\mathop {\lim }\limits_{x \to {t^ + }} a(x) = {A_1} {/eq} and {eq}\mathop {\lim }\limits_{x \to {t^ - }} a(x) = {A_2} {/eq}, then;

(1) Limit exist if,

{eq}{A_1} = {A_2} {/eq}

(2) Limit does not exist if,

{eq}{A_1} \ne {A_2} {/eq}

Answer and Explanation:

{eq}\displaystyle \eqalign{ & \mathop {\lim }\limits_{t \to 0} \left( {\frac{{sin(5t)}}{{sin(4t)}}i + \frac{{\ln (\sin (5t))}}{{\ln (\tan (4t))}}j + t\ln (5t)k} \right) \cr & \left( {\mathop {\lim }\limits_{t \to 0} \frac{{sin(5t)}}{{sin(4t)}}i + \mathop {\lim }\limits_{t \to 0} \frac{{\ln (\sin (5t))}}{{\ln (\tan (4t))}}j + \mathop {\lim }\limits_{t \to 0} t\ln (5t)k} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From limit properties}}} \right) \cr & \mathop {\lim }\limits_{t \to 0} \frac{{sin(5t)}}{{sin(4t)}} \cr & \frac{5}{4}\mathop {\lim }\limits_{t \to 0} \frac{{\sin 5t}}{{5t}}\mathop {\lim }\limits_{t \to 0} \frac{{4t}}{{\sin 4t}} \cr & \frac{5}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right) \cr & \cr & \mathop {\lim }\limits_{t \to 0} \frac{{\ln (\sin (5t))}}{{\ln (\tan (4t))}} \cr & \frac{{\ln \left( {\mathop {\lim }\limits_{t \to 0} 5t\mathop {\lim }\limits_{t \to 0} \frac{{\sin (5t)}}{{5t}}} \right)}}{{\ln \left( {\mathop {\lim }\limits_{t \to 0} 4t\mathop {\lim }\limits_{t \to 0} \frac{{\tan (4t)}}{{4t}}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From limit properties}}} \right) \cr & \frac{{\ln \left( 0 \right)}}{{\ln \left( 0 \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right) \cr & \frac{{ - \infty }}{{ - \infty }} \cr & {\text{undefine}} \cr & \cr & \mathop {\lim }\limits_{t \to 0} t\ln (5t) \cr & \mathop {\lim }\limits_{t \to 0} t\mathop {\lim }\limits_{t \to 0} \ln (5t)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From limit properties}}} \right) \cr & 0 \cr & \cr & \mathop {\lim }\limits_{t \to 0} \left( {\frac{{sin(5t)}}{{sin(4t)}}i + \frac{{\ln (\sin (5t))}}{{\ln (\tan (4t))}}j + t\ln (5t)k} \right) = \left( {\frac{5}{4}i + \left( 0 \right)k} \right) \cr & \cr & \cr & \mathop {\lim }\limits_{t \to 5} \left( {\left( {\frac{{{t^3} - 125}}{{t - 5}}} \right)i + \frac{{{t^2} - 9t + 20}}{{{t^2} - 3t - 10}}j + (t - 5)ln(t - 5)k} \right) \cr & \left( {\mathop {\lim }\limits_{t \to 5} \left( {\frac{{{t^3} - 125}}{{t - 5}}} \right)i + \mathop {\lim }\limits_{t \to 5} \frac{{{t^2} - 9t + 20}}{{{t^2} - 3t - 10}}j + \mathop {\lim }\limits_{t \to 5} (t - 5)ln(t - 5)k} \right) \cr & \mathop {\lim }\limits_{t \to 5} \left( {\frac{{{t^3} - 125}}{{t - 5}}} \right) \cr & 3{\left( 5 \right)^{3 - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\mathop {\lim }\limits_{x \to a} \left( {\frac{{{x^n} - {a^n}}}{{x - a}}} \right) = n{a^{n - 1}}} \right) \cr & 3(25) \cr & 75 \cr & \cr & \mathop {\lim }\limits_{t \to 5} \frac{{{t^2} - 9t + 20}}{{{t^2} - 3t - 10}} \cr & \mathop {\lim }\limits_{t \to 5} \frac{{\left( {t - 4} \right)\left( {t - 5} \right)}}{{\left( {t + 2} \right)\left( {t - 5} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{t^2} - 9t + 20 = \left( {t - 4} \right)\left( {t - 5} \right),{t^2} - 3t - 10 = \left( {t + 2} \right)\left( {t - 5} \right)} \right) \cr & \mathop {\lim }\limits_{t \to 5} \frac{{\left( {t - 4} \right)}}{{\left( {t + 2} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Cancel the common term}}} \right) \cr & \frac{{\left( {5 - 4} \right)}}{{\left( {5 + 2} \right)}} \cr & \frac{1}{7} \cr & \cr & \mathop {\lim }\limits_{t \to 5} (t - 5)\ln (t - 5) \cr & \mathop {\lim }\limits_{t \to 5} (t - 5)\mathop {\lim }\limits_{t \to 5} \ln (t - 5)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From limit properties}}} \right) \cr & 0 \cr & \cr & \mathop {\lim }\limits_{t \to 5} \left( {\left( {\frac{{{t^3} - 125}}{{t - 5}}} \right)i + \frac{{{t^2} - 9t + 20}}{{{t^2} - 3t - 10}}j + (t - 5)ln(t - 5)k} \right) = \left( {75i + \frac{1}{7}j + 0k} \right) \cr} {/eq}


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Understanding the Properties of Limits

from Math 104: Calculus

Chapter 6 / Lesson 5
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