Evaluate the following limit : lim(x→1/2) (8x^3 + 1)/(2x + 1)

Question

Evaluate the following limit : \(\lim\limits_{\text x \to 1/2}\cfrac{8\text x^3+1}{2\text x+1} \)

lim(x→1/2) (8x³ + 1)/(2x + 1)

Answer 1

We need to find the limit for : \(\lim\limits_{\text x \to 1/2}\cfrac{8\text x^3+1}{2\text x+1} \)

As limit can’t be find out simply by putting x = (–1/2) because it is taking indeterminate form(0/0) form, so we need to have a different approach.

Let, Z = \(\lim\limits_{\text x \to 1/2}\cfrac{8\text x^3+1}{2\text x+1} \)

 \(\Rightarrow\) Z = \(\lim\limits_{\text x \to 1/2}\cfrac{(2\text x)^3-(-1)}{2\text x-(-1)} \)

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 = \(\lim\limits_{\text x \to 1/2}\cfrac{(2\text x)^3-(-1)}{2\text x-(-1)} \)

Let y = 2x

As x → –1/2 ⇒ 2x = y → –1

 \(\therefore\) Z

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

 \(\therefore\) Z = 3(-1)^{3 - 1} = 3(-1)² = 3

Hence, \(\lim\limits_{\text x \to 1/2}\cfrac{8\text x^3+1}{2\text x+1} \) = 3