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Question: How do you evaluate the limit \(\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}\) as \(h \to 0\)?...

How do you evaluate the limit (3+h)3−27h\dfrac{{{\left( 3+h \right)}^{3}}-27}{h} as h→0h \to 0?

Explanation

Solution

We will apply the algebraic formula (a+b)3=a3+b3+3a2b+3ab2{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}} and solve the function to get the desired limit. Since, the limit is zero and we have only one h in the denominator so, we will not use direct substitution here. We will try to eliminate the denominator in order to get the answer. And this can be done by using the algebraic formula.

Complete step-by-step answer:
By the term limit we mean the nearness of the function with its limit. To know how near is the limit of the function (3+h)3−27h\dfrac{{{\left( 3+h \right)}^{3}}-27}{h} as h approaches to the point 0, we will use algebraic cubic formula. This formula is (a+b)3=a3+b3+3a2b+3ab2{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}. So, we can write,
(3+h)3=(3)3+(h)3+3(3)2(h)+3(3)(h)2 ⇒(3+h)3=27+h3+27h+9h2    ...(i) \begin{aligned} & {{\left( 3+h \right)}^{3}}={{\left( 3 \right)}^{3}}+{{\left( h \right)}^{3}}+3{{\left( 3 \right)}^{2}}\left( h \right)+3\left( 3 \right){{\left( h \right)}^{2}} \\\ & \Rightarrow {{\left( 3+h \right)}^{3}}=27+{{h}^{3}}+27h+9{{h}^{2}}\,\,\,\,...(i) \\\ \end{aligned}
Now we will consider lim⁡h→0(3+h)3−27h\displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}and substitute the value of (i) in this function. Therefore, we get
lim⁡h→0(3+h)3−27h=lim⁡h→027+h3+27h+9h2−27h ⇒lim⁡h→0(3+h)3−27h=lim⁡h→027+h3+27h+9h2−27h ⇒lim⁡h→0(3+h)3−27h=lim⁡h→0h3+27h+9h2h \begin{aligned} & \displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}=\displaystyle \lim_{h \to 0}\dfrac{27+{{h}^{3}}+27h+9{{h}^{2}}-27}{h} \\\ & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}=\displaystyle \lim_{h \to 0}\dfrac{{27}+{{h}^{3}}+27h+9{{h}^{2}}-{27}}{h} \\\ & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}=\displaystyle \lim_{h \to 0}\dfrac{{{h}^{3}}+27h+9{{h}^{2}}}{h} \\\ \end{aligned}
At this step we will take h as common. Thus, we get

& \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}=\displaystyle \lim_{h \to 0}\dfrac{h\left( {{h}^{2}}+27+9h \right)}{h} \\\ & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}=\displaystyle \lim_{h \to 0}\left( {{h}^{2}}+27+9h \right)\,\,...(ii) \\\ \end{aligned}$$ Now, it is easy for us to substitute the value of h = 0 in equation (ii). This results into the following, $$\begin{aligned} & \displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}=\displaystyle \lim_{h \to 0}\left( {{h}^{2}}+27+9h \right) \\\ & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}={{\left( 0 \right)}^{2}}+27+9\left( 0 \right) \\\ & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}=27 \\\ \end{aligned}$$ Hence, the value of the limit $$\displaystyle \lim_{h \to 0}\dfrac{{{\left( 3+h \right)}^{3}}-27}{h}=27$$. **Note:** Since, the denominator is h here and the limit is 0 so, it is clear that we cannot put a direct value of h. Otherwise, the limit will be not defined which we do not want at all. So it will be nice if we delete the denominator somehow. We can do this by using an algebraic formula in numerator. By converting ${{\left( 3+h \right)}^{3}}=27+{{h}^{3}}+27h+9{{h}^{2}}$ we come closer to deleting the constant which will further cancels one h from numerator and denominator both. After this the path becomes clear as to substitute the value of h in it and get the desired value. In case, we directly put h = 0 in the function at first and write the answer is not defined then it will become the wrong answer. So, it is important to focus on this point in order to get the right answer.