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Question: How do you find the limit of \[\dfrac{{{5^t} - {3^t}}}{t}\] as x approaches \[0\] ?...

How do you find the limit of 5t−3tt\dfrac{{{5^t} - {3^t}}}{t} as x approaches 00 ?

Explanation

Solution

Here in this question, we need to find the limit for the function. Let we define the given function as f(t)f(t) and we apply the limit to it. The function will get closer and closer to some number. When the x approaches to 00 we need to find the limit of a function.

Complete step by step answer:
The idea of a limit is a basis of all calculus. The limit of a function is defined as let f(t)f(t) be a function defined on an interval that contains t=at = a. Then we say that lim⁡t→af(t)=L\mathop {\lim }\limits_{t \to a} f(t) = L, if for every ε>0\varepsilon > 0there is some number δ>0\delta > 0 such that ∣f(t)−L∣<ε\left| {f(t) - L} \right| < \varepsilon whenever0<∣t−a∣<δ0 < \left| {t - a} \right| < \delta .
Here we have to find the value of the limit when x is zero. Let we define the given function as f(t)=5t−3ttf(t) = \dfrac{{{5^t} - {3^t}}}{t}. Now we are going to apply the limit to the function f(t)f(t)so we have
lim⁡t→0f(t)=lim⁡t→05t−3tt\mathop {\lim }\limits_{t \to 0} f(t) = \mathop {\lim }\limits_{t \to 0} \dfrac{{{5^t} - {3^t}}}{t}
In the numerator add and subtract 1, we get
lim⁡t→0f(t)=lim⁡t→05t−1−3t+1t\mathop {\lim }\limits_{t \to 0} f(t) = \mathop {\lim }\limits_{t \to 0} \dfrac{{{5^t} - 1 - {3^t} + 1}}{t}
This can be written as
lim⁡t→0f(t)=lim⁡t→05t−1−(3t−1)t\mathop {\lim }\limits_{t \to 0} f(t) = \mathop {\lim }\limits_{t \to 0} \dfrac{{{5^t} - 1 - ({3^t} - 1)}}{t}
The denominator is applied to all terms, so this can be written as
lim⁡t→0f(t)=lim⁡t→05t−1t−(3t−1)t\mathop {\lim }\limits_{t \to 0} f(t) = \mathop {\lim }\limits_{t \to 0} \dfrac{{{5^t} - 1}}{t} - \dfrac{{({3^t} - 1)}}{t}
The limit is applied to both the terms
lim⁡t→0f(t)=lim⁡t→05t−1t−lim⁡t→0(3t−1)t\mathop {\lim }\limits_{t \to 0} f(t) = \mathop {\lim }\limits_{t \to 0} \dfrac{{{5^t} - 1}}{t} - \mathop {\lim }\limits_{t \to 0} \dfrac{{({3^t} - 1)}}{t}
By the formula we know that the lim⁡h→0ah−1h=ln⁡ a  ; (a>0)\mathop {\lim }\limits_{h \to 0} \dfrac{{{a^h} - 1}}{h} = \ln \,a\,\,;\,(a > 0)
The function f(t)f(t) is an exponential function. As x approaches to 00 we have to substitute the t as00.
⇒lim⁡t→0f(t)=ln⁡ 5 − ln⁡ 3\Rightarrow \mathop {\lim }\limits_{t \to 0} f(t) = \ln \,5\, - \,\ln \,3
By using the property of the logarithmic functions the above inequality is written as
⇒lim⁡t→0f(t)=ln⁡ 53\Rightarrow \mathop {\lim }\limits_{t \to 0} f(t) = \ln \,\dfrac{5}{3}
Therefore we have found the limit for the function f(t)=5t−3ttf(t) = \dfrac{{{5^t} - {3^t}}}{t}
Hence lim⁡t→05t−3tt=ln⁡53\mathop {\lim }\limits_{t \to 0} \dfrac{{{5^t} - {3^t}}}{t} = \ln \dfrac{5}{3}

Note: We have some standard formulas for the limit, we should be aware of it while solving the problems. We got the solution in the form of logarithm so we have to know about the properties of the logarithmic function and mainly we should know about the definition of the limit function.