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Question: Find the following limit \(\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + ...

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

Explanation

Solution

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

Complete step by step solution:
lim⁡x→0ex−e−x−2log⁡(1+x)x sin⁡x\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,
=l⁡tx→0ex−e−x−2log⁡(1+x)x sin⁡x×xx= \mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\,\sin x}} \times \dfrac{x}{x}
=l⁡tx→0ex−e−x−2log⁡(1+x)x2 sin⁡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
=l⁡t        x→0ex−e−x−2log⁡(1+x)x2 ×xsin⁡ 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
l⁡tx→0sin⁡x=1\mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{\sin }}{x} = 1
l⁡tx→0ex−e−x−2log⁡(1+x)x2 .1\mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{{x^2}\,}}.1
l⁡tx→0ex−e−x−2log⁡(1+x)x2 \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 (00)\left( {\dfrac{0}{0}} \right)
So, use L’ hospital’s Rule
l⁡tx→0ex+e−x−21+x2x               [∵(log⁡x)′=1x]\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 (00)\left( {\dfrac{0}{0}} \right)
Again applying L’ Hospital’s Rule
l⁡tx→0ex−e−x+2(1+x)22 \mathop {\operatorname{l} t}\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} + \dfrac{2}{{{{(1 + x)}^2}}}}}{2}\,
On substitution the value of l⁡tx→0\mathop {\operatorname{l} t}\limits_{x \to 0} in this
=1−1+212 =22 =1  = \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 (00)\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.