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Question: How do you find the limit \[\displaystyle \lim_{h \to 0}\dfrac{{{\left( 2+h \right)}^{3}}-8}{h}\]?...

How do you find the limit lim⁡h→0(2+h)3−8h\displaystyle \lim_{h \to 0}\dfrac{{{\left( 2+h \right)}^{3}}-8}{h}?

Explanation

Solution

In the given question, we are given a question based on limits. We are given an expression and find the limit of the given expression. We will not directly substitute the value of ‘h’, which is 0, in the expression because then we will get an indeterminate value 00\dfrac{0}{0}. We will proceed by solving the given expression first. We will use the algebraic identity, (a+b)3=a3+b3+3a2b+3ab2{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}} in the given expression and reduce the expression as much as possible. Hence, we will get the limit of the expression.

Complete step by step solution:
According to the given question, we are given an expression which we have to evaluate and find the limit. So, the given expression we have is,
lim⁡h→0(2+h)3−8h\displaystyle \lim_{h \to 0}\dfrac{{{\left( 2+h \right)}^{3}}-8}{h}
Here, we have to find the value of the expression when h→0h \to 0.
We will not directly substitute the value of ‘h’ as 0, because then we will get a value in an indeterminate form, 00\dfrac{0}{0}. So, we will first solve the expression.
We will use the possible identities to reduce the expression as much as possible and only then we will put the value of ‘h’ as 0.
We will first use the algebraic identity, (a+b)3=a3+b3+3a2b+3ab2{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}, to expand the expression. We get,
⇒lim⁡h→08+h3+3(2)2h+6h2−8h\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{8+{{h}^{3}}+3{{\left( 2 \right)}^{2}}h+6{{h}^{2}}-8}{h}
Cancelling the similar terms, we get,
=lim⁡h→0h3+12h+6h2h= \displaystyle \lim_{h \to 0}\dfrac{{{h}^{3}}+12h+6{{h}^{2}}}{h}
From the above expression, we can take ‘h’ as common and so we get,
=lim⁡h→0h(h2+12+6h)h= \displaystyle \lim_{h \to 0}\dfrac{h\left( {{h}^{2}}+12+6h \right)}{h}
We will now cancel ‘h’ from both numerator and denominator and we get,
=lim⁡h→0h2+12+6h=\displaystyle \lim_{h \to 0}{{h}^{2}}+12+6h
Applying the limits now, we will now get the value of the expression as,
=0+12+6(0)= 0+12+6\left( 0 \right)
=12= 12
Therefore, the limit of the expression is 12.

Note: If we had applied the limits in the beginning then we would not have got this value, we would have been still stuck with the 00\dfrac{0}{0}, which is a wrong answer. We solve these expressions in order to get rid of these indeterminate forms. So, if we encounter an indeterminate value on direct substitution, we just need to solve the expression and reduce it and only then apply the limits.