Evaluate \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x}\] .
Answer
589.8k+ views
Hint: We are asked to evaluate a given function in the limit \[x = 0\] . First check the value just by putting \[x = 0\] in the function and if it tends to infinity, then go for the L’Hospital’s rule. Recall the formula for L’Hospital’s rule and use this to evaluate the given function.
Complete step-by-step answer:
We are asked to solve \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x}\]
We observe that if we put \[x = 0\] in the given function, then the function will be \[f(x) = \dfrac{0}{0} \]
So, we apply L’Hospital’s rule.
According to L’Hospital’s rule, we have
\[\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}\] (i)
Comparing with the given function, here \[f(x) = {(1 - x)^n} - 1\] and \[g(x) = x\] (ii)
Therefore, \[f'(x) = \dfrac{d}{{dx}}\left[ {{{(1 - x)}^n} - 1} \right] \]
\[ \Rightarrow f'(x) = n{(1 - x)^{n - 1}}( - 1)\]
\[ \Rightarrow f'(x) = - n{(1 - x)^{n - 1}}\] (iii)
And \[g'(x) = \dfrac{d}{{dx}}\left( x \right)\]
\[ \Rightarrow g'(x) = 1\] (iv)
Applying L’Hospital’s rule and using equations (ii), (iii) and (iv) in equation (i) we get,
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - n{{(1 - x)}^{n - 1}}}}{1}\]
Now putting \[x = 0\] we get,
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x} = \dfrac{{ - n{{(1 - 0)}^{n - 1}}}}{1}\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x} = - n\]
Therefore, the required answer is \[ - n\] .
So, the correct answer is “\[ - n\]”.
Note: Whenever you are asked to find a function in a limit, first apply the limit and check if it gives you a finite answer and if not then try to simplify the function such that when you apply the limit you don’t get the denominator as zero and get a finite answer. And if it is difficult to simplify the given function then go for L’Hospital rule. Always remember the formula for L’Hospital’s rule as it helps to make the problem easier to solve. All we need to do in this rule is to differentiate the numerator and denominator as we have done for this problem and take the limit and solve it. And one more important point that one should remember is L’Hospital’s rule can only be applied when the function tends to infinity after taking the limit.
Complete step-by-step answer:
We are asked to solve \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x}\]
We observe that if we put \[x = 0\] in the given function, then the function will be \[f(x) = \dfrac{0}{0} \]
So, we apply L’Hospital’s rule.
According to L’Hospital’s rule, we have
\[\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}\] (i)
Comparing with the given function, here \[f(x) = {(1 - x)^n} - 1\] and \[g(x) = x\] (ii)
Therefore, \[f'(x) = \dfrac{d}{{dx}}\left[ {{{(1 - x)}^n} - 1} \right] \]
\[ \Rightarrow f'(x) = n{(1 - x)^{n - 1}}( - 1)\]
\[ \Rightarrow f'(x) = - n{(1 - x)^{n - 1}}\] (iii)
And \[g'(x) = \dfrac{d}{{dx}}\left( x \right)\]
\[ \Rightarrow g'(x) = 1\] (iv)
Applying L’Hospital’s rule and using equations (ii), (iii) and (iv) in equation (i) we get,
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - n{{(1 - x)}^{n - 1}}}}{1}\]
Now putting \[x = 0\] we get,
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x} = \dfrac{{ - n{{(1 - 0)}^{n - 1}}}}{1}\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - x)}^n} - 1}}{x} = - n\]
Therefore, the required answer is \[ - n\] .
So, the correct answer is “\[ - n\]”.
Note: Whenever you are asked to find a function in a limit, first apply the limit and check if it gives you a finite answer and if not then try to simplify the function such that when you apply the limit you don’t get the denominator as zero and get a finite answer. And if it is difficult to simplify the given function then go for L’Hospital rule. Always remember the formula for L’Hospital’s rule as it helps to make the problem easier to solve. All we need to do in this rule is to differentiate the numerator and denominator as we have done for this problem and take the limit and solve it. And one more important point that one should remember is L’Hospital’s rule can only be applied when the function tends to infinity after taking the limit.
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