Given, \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}=\lim\limits_{\text x \to 5} \)(4 + x) 

we need to find value of n

So we will first find the limit and then equate it with \(\lim\limits_{\text x \to5}\)(4 + x) to get the value of n.

We need to find the limit for : \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}\)

As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.

Let, Z = \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}\)

Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.

Formula to be used: \(\lim\limits_{\text x \to a}\cfrac{(\text x)^n-(a)^n}{\text x-a} \) = na^{n -1}

As Z matches exactly with the form as described above so we don’t need to do any manipulations–

Z = \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}\)

Use the formula: \(\lim\limits_{\text x \to a}\cfrac{(\text x)^n-(a)^n}{\text x-a} \) = na^{n -1}

∴ Z = 9(a)^9–1 = 9a⁸

According to question Z = \(\lim\limits_{\text x \to5}\)(4 + x) = 4 + 5 = 9

∴ 9(a)⁸ = 9

⇒ a⁸ = 1 = 1 8 or (–1)8

Clearly on comparison we have –

a = 1 or –1