Question

Lim x^5-a^5/x^4-a^4
x-> a

Answer 1

Calculus - Limits

We are asked to evaluate the given limits:

$\longrightarrow \lim \limits_{x \to a} \dfrac{x^5 - a^5}{x^4 - a^4}$

By directly substituting the limits x → a, we obtain:

$\dfrac{a^5 - a^5}{a^4 - a^4} = \dfrac{a - a}{a - a} = \boxed{\dfrac{0}{0}}$

Which is an indeterminate quantity, therefore substitution method failed to evaluate our limit. So we have to choose another method to solve the problem.

Solution:

We can see that factorisation and rationalization is not possible. So let's solve this problem by applying the L'Hospital's rule.

L'Hospital's rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

Therefore, taking derivative of the numerator [top part of the fraction] and denominator [bottom part of the fraction], then evaluating the limits, we obtain:

$\implies \lim \limits_{x \to a}\Bigg\{\dfrac{\frac{d}{dx}(x^5 - a^5)}{\frac{d}{dx}(x^4 - a^4)} \Bigg\}$

$\implies \lim \limits_{x \to a}\Bigg\{\dfrac{5x^{5 - 1} - 0}{4x^{4 - 1} - 0} \Bigg\}$

$\implies \lim \limits_{x \to a}\Bigg\{\dfrac{5x^{4} - 0}{4x^{3} - 0} \Bigg\}$

$\implies \lim \limits_{x \to a}\Bigg\{\dfrac{5x^{4}}{4x^{3}} \Bigg\}$

$\implies \lim \limits_{x \to a}\Bigg\{\dfrac{5x}{4} \Bigg\}$

Now evaluate the limit of 'x' by plugging in 'a' for 'x'.

$\implies \dfrac{5a}{4}$

Therefore the required answer is:

$\boxed{\lim \limits_{x \to a} \dfrac{x^5 - a^5}{x^4 - a^4} = \dfrac{5a}{4}}$

$\rule{300}{2}$

Formula used:

The following are the formulas that have been used to find the solution:

$\begin{gathered}\boxed{\begin{array}{l}\bullet \;\;\dfrac{d}{dx}(x^n) = nx^{n-1} \\ \\ \bullet \;\;\dfrac{d}{dx}(\rm{constant}) = 0\end{array}}\end{gathered}$

To read similar type of question from this chapter, refer the below link:

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Answer 2

Answer:

5a/4

Step-by-step explanation:

We need to evaluate the given limit.

$\longrightarrow \sf \lim \limits_{x \to a}\left( \dfrac{ {x}^{5} - {a}^{5} }{ {x}^{4} - {a}^{4} }\right)$

By directly substituting the limiting value, we get 0/0 which is meaningless as it is an indeterminate quantity. Therefore we have to do more work to solve the problem.

We are aware about below formula:

• $\boxed{\sf \lim \limits_{x \to a}\dfrac{ {x}^{n} - {a}^{n} }{ x - a } = n(a)^{n - 1} }$

We can see that the given question is somewhat relating to this identity. The only requirment is to have (x - a) in denominator and (x - a) in numerator to apply the identity.

Multiply both numerator and denominator with (x - a)

$\longrightarrow \sf \lim \limits_{x \to a}\left( \dfrac{ {x}^{5} - {a}^{5} }{ {x}^{4} - {a}^{4} } \times \dfrac{x - a}{x - a} \right)$

$\longrightarrow \sf \lim \limits_{x \to a}\left( \dfrac{ {x}^{5} - {a}^{5} }{x - a} \times \dfrac{x - a}{ {x}^{4} - {a}^{4} } \right)$

${\longrightarrow \sf \lim \limits_{x \to a}\left( \dfrac{ {x}^{5} - {a}^{5} }{x - a}\right) \times \lim \limits_{x \to a}\left( \dfrac{ {x}^{4} - {a}^{4} }{x - a} \right)^{ - 1}}$

Now make use of the identity.

${\longrightarrow \sf5a^{4} \times\left(4 {a}^{3} \right)^{ - 1}}$

${\longrightarrow \sf5a^{4} \times\left( \dfrac{1}{4 {a}^{3} }\right)}$

${\longrightarrow \sf \dfrac{5a}{4}}$

So our required answer is:

$\orange{\boxed{\sf \lim \limits_{x \to a}\left( \dfrac{ {x}^{5} - {a}^{5} }{ {x}^{4} - {a}^{4} }\right) = \frac{5a}{4} }}$

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