Question

Find the following limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\,\sin x}}$

Answer 1

Hint: Use L’ Hospital’s Rule to find out the solution and also use formula of limit for trigonometric function $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin }}{x} = 1$,


Complete step by step solution:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\,\sin x}}$
Now, solve denominator term by multiplying x in numerator and denominator,
$ = \mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\,\sin x}} \times \dfrac{x}{x}$
$ = \mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{{x^2}\,\sin x}} \times x$
$\mathop { = \operatorname{l} t}\limits_{\,\,\,\,\,\,\,\,x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{{x^2}\,}} \times \dfrac{x}{{\sin \,x}}$
Since, we know
$\mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{\sin }}{x} = 1$
$\mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{{x^2}\,}}.1$
$\mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{{x^2}\,}}$
On applying the limit, we see that it is of the form $\left( {\dfrac{0}{0}} \right)$
So, use L’ hospital’s Rule
$\mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} + {e^{ - x}} - \dfrac{2}{{1 + x}}}}{{2x\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\because (\log x)' = \dfrac{1}{x}]$
Let us check we see that it is of the form $\left( {\dfrac{0}{0}} \right)$
Again applying L’ Hospital’s Rule
$\mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} + \dfrac{2}{{{{(1 + x)}^2}}}}}{2}\,$
On substitution the value of $\mathop {\operatorname{l} t}\limits_{x \to 0} $in this
$
   = \dfrac{{1 - 1 + \dfrac{2}{1}}}{2} \\
   = \dfrac{2}{2} \\
   = 1 \\
 $


Note: Students must check the type of question which is given, whether it is of the form $\left( {\dfrac{0}{0}} \right)$or $\left( {\dfrac{\infty }{\infty }} \right)$after applying limits, then apply L’ Hospital’s Rule till we get non zero number.