Question

Find the limit.

Answer 1

Step-by-step explanation:

Solution−

Given expression is ,

$\large{lim_{x→5} \: \frac{ {x}^{2} - 5 }{ {x}^{2} + x - 30} }$

if we substitute directly x=5, we get

$\longrightarrow \: \frac{ {5}^{2} - 5}{ {5}^{2} + 5 - 30 }$

$\longrightarrow \: \frac{25 - 5}{30 - 30}$

$\longrightarrow \: \frac{20}{0}$

$\longrightarrow \: \infty$

Hence,

${\red{ \boxed{\large{lim_{x→5} \: \frac{ {x}^{2} - 5 }{ {x}^{2} + x - 30 } = \infty}}}}$

Answer 2

Answer:

DNE (Doesn’t exist)

Step-by-step explanation:

Hi! We are given the limit expression:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-5}{x^2+x-30}}$

First step of evaluating limit is always directly substitution - substitute x = 5 in the expression.

$\displaystyle \large{ \lim_{x\to 5}\frac{5^2-5}{5^2+5-30}}\\ \displaystyle \large{ \lim_{x\to 5}\frac{25-5}{25+5-30}}\\ \displaystyle \large{ \lim_{x\to 5}\frac{20}{0}}$

Looks like we’ve got 20/0 after direct substitution. Note that this isn’t an indeterminate form but undefined since it’s not 0/0 which would make things different.

Now, plot the graph and see at x approaches 5, the function y approaches both positive infinity and negative infinity.

Introducing, left limit would be:

$\displaystyle \large{ \lim_{x \to 5^{-}} \frac{x^2-5}{x^2+x-30} = -\infty}$

And right limit is:

$\displaystyle \large{ \lim_{x \to 5^{+}} \frac{x^2-5}{x^2+x-30} = \infty}$

In limit, if both left and right limit are not equal then the limit does not exist.

From left limit and right, both are not equal. Therefore, the limit does not exist.

Cautions:

• Limit being DNE in this case simply means both left and right limit are not equal.
• The limit does not approach infinity a left limit apporoaches negative infinity as well which is different from positive infinity hence limit being DNE.