Question

$\sf \: Find \: the \: general \: limit \: of \: the \: following \: function : \\ \sf \: lim_{x \to \: 9} (x² + 2^7 + (9.1 × 10))$​

Answer 1

Solution:-

$9( {x}^{2} + {2}^{7} + (9.1 \times 10)$

$⟹ \: 9( {x}^{2} + {2}^{7} + 91.0)$

$⟹ \: 9 \times {x}^{2} + 9 \times {2}^{7} + 9 \times 91$

$⟹ \: 9 {x}^{2} + 9 \times 128 + 819$

$⟹ \: 9 {x}^{2} + 1152 + 819$

$⟹ \: 9 {x}^{2} + 1971$

Hope it helps!

Answer 2

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

Given expression is

$\rm :\longmapsto\: \displaystyle \lim_{x \to \: 9} [ {x}^{2} + 2^7 + (9.1 × 10)]$

Let us first evaluate the Left Hand Limit

$\rm :\longmapsto\: \displaystyle \lim_{x \to \: 9^ - } [ {x}^{2} + 2^7 + (9.1 × 10)]$

To evaluate this, we have to use method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = 9 - h \: \: as \: x \to \: 9 \: \: so \: h \to \: 0}$

So, above expression can be rewritten as

$\rm \: = \: lim_{h \to \: 0} \bigg[ {(9 - h)}^{2} + 128 + 91\bigg]$

$\rm \: = \: {(9 - 0)}^{2} + 219$

$\rm \: = \: 81 + 219$

$\rm \: = \: 300$

$\bf\implies \: \displaystyle \lim_{x \to \: 9^ - } \bf [{x}^{2} + 2^7 + (9.1 × 10)]= 300$

Now, Consider Right Hand Limit

$\rm :\longmapsto\: \displaystyle \lim_{x \to \: 9^ + } [ {x}^{2} + 2^7 + (9.1 × 10)]$

To evaluate this limit, we use method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = 9 + h \: \: as \: x \to \: 9 \: \: so \: h \to \: 0}$

So, above expression can be rewritten as

$\rm \: = \: lim_{h \to \: 0} \bigg[ {(9 + h)}^{2} + 128 + 91\bigg]$

$\rm \: = \: {(9 + 0)}^{2} + 219$

$\rm \: = \: 81 + 219$

$\rm \: = \: 300$

$\bf\implies \: \displaystyle \lim_{x \to \: 9^ + } \bf [{x}^{2} + 2^7 + (9.1 × 10)]= 300$

So, we concluded that

$\boxed{\tt{ \displaystyle \lim_{x \to 9^ + } \sf [{x}^{2} + 2^7 + (9.1 × 10)] = \lim_{x \to 9^ - } \sf [{x}^{2} + 2^7 + (9.1 × 10)]}}$

Hence,

$\sf\implies\boxed{\tt{ \displaystyle \lim_{x \to 9} \sf [{x}^{2} + 2^7 + (9.1 × 10)] = 300}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

EXPLORE MORE

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

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

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

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

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