# Evaluate the Following Limit: Lim H → 0 ( a + H ) 2 Sin ( a + H ) − a 2 Sin a H - Mathematics

Evaluate the following limit:

$\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}$

#### Solution

$\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}$
$= \lim_{h \to 0} \frac{\left( a^2 + 2ah + h^2 \right)\sin\left( a + h \right) - a^2 \sin a}{h}$
$= \lim_{h \to 0} \frac{\left( 2ah + h^2 \right)\sin\left( a + h \right) + a^2 \sin\left( a + h \right) - a^2 \sin a}{h}$
$= \lim_{h \to 0} \frac{\left( 2ah + h^2 \right)\sin\left( a + h \right)}{h} + \lim_{h \to 0} \frac{a^2 \sin\left( a + h \right) - a^2 \sin a}{h}$
$= \lim_{h \to 0} \left( 2a + h \right)\sin\left( a + h \right) + a^2 \lim_{h \to 0} \frac{\sin\left( a + h \right) - \sin a}{h}$
$\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}$
$= \lim_{h \to 0} \frac{\left( a^2 + 2ah + h^2 \right)\sin\left( a + h \right) - a^2 \sin a}{h}$
$= \lim_{h \to 0} \frac{\left( 2ah + h^2 \right)\sin\left( a + h \right) + a^2 \sin\left( a + h \right) - a^2 \sin a}{h}$
$= \lim_{h \to 0} \frac{\left( 2ah + h^2 \right)\sin\left( a + h \right)}{h} + \lim_{h \to 0} \frac{a^2 \sin\left( a + h \right) - a^2 \sin a}{h}$
$= \lim_{h \to 0} \left( 2a + h \right)\sin\left( a + h \right) + a^2 \lim_{h \to 0} \frac{\sin\left( a + h \right) - \sin a}{h}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.7 | Q 62 | Page 51