Question

Simplify the following limit

$\lim \: x \: \to \: 2 \: \frac{x + {x}^{2} + {x}^{3} + {x}^{4} - 30 }{x - 2}$

Don't apply L Hospital Rule​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{x \to 2}\rm \frac{x + {x}^{2} + {x}^{3} + {x}^{4} - 30}{x - 2}$

If we substitute directly x = 2, we get

$\rm \: = \frac{2 + {2}^{2} + {2}^{3} + {2}^{4} - 30}{2 - 2}$

$\rm \: = \frac{2 + 4 + 8 + 16- 30}{0}$

$\rm \: = \frac{30- 30}{0}$

$\rm \: = \frac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x \to 2}\rm \frac{x + {x}^{2} + {x}^{3} + {x}^{4} - 30}{x - 2}$

can be rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{x + {x}^{2} + {x}^{3} + {x}^{4} - (2 + 4 + 8 + 16)}{x - 2}$

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{x + {x}^{2} + {x}^{3} + {x}^{4} - 2 - 4 - 8 - 16}{x - 2}$

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{(x - 2) + ({x}^{2} - 4) + ({x}^{3} - 8) + ({x}^{4} - 16)}{x - 2}$

can be further rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{x - 2}{x - 2} + \displaystyle\lim_{x \to 2}\rm \frac{ {x}^{2} - {2}^{2} }{x - 2} + \displaystyle\lim_{x \to 2}\rm \frac{ {x}^{3} - {2}^{3} }{x - 2} + \displaystyle\lim_{x \to 2}\rm \frac{ {x}^{4} - {2}^{4} }{x - 2}$

We know,

$\boxed{\tt{ \displaystyle\lim_{x \to a}\rm \frac{ {x}^{n} - {a}^{n} }{x - a} \: = {na}^{n - 1} \: }}$

So, using this result, we get

$\rm \: = \: 1 + {2(2)}^{2 - 1} + {3(2)}^{2 - 1} + {4(2)}^{4 - 1}$

$\rm \: = \: 1 + 4 + 12 + 32$

$\rm \: = \: 49$

Hence,

$\rm\implies \:\boxed{\tt{ \displaystyle\lim_{x \to 2}\rm \frac{x + {x}^{2} + {x}^{3} + {x}^{4} - 30}{x - 2} = 49 \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} \: = \: 1$

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} \: = \: 1$

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} \: = \: 1$

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1 }{x} \: = \: 1$

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1 }{x} \: = \: loga$