Question Number 200298 by Calculusboy last updated on 16/Nov/23
Commented by Frix last updated on 17/Nov/23
$$−\frac{\mathrm{1}}{\mathrm{16}} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Nov/23
$$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\:\sqrt{{x}}\:}−\frac{\mathrm{1}}{\mathrm{2}}}{{x}−\mathrm{4}} \\ $$$$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}−\sqrt{{x}}}{\:\mathrm{2}\sqrt{{x}}\:}}{\:\left(\sqrt{{x}}\:−\mathrm{2}\right)\left(\sqrt{{x}}\:+\mathrm{2}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{−\cancel{\left(\sqrt{{x}}\:−\mathrm{2}\right)}}{\:\mathrm{2}\sqrt{{x}}\:\cancel{\left(\sqrt{{x}}\:−\mathrm{2}\right)}\left(\sqrt{{x}}\:+\mathrm{2}\right)} \\ $$$$=\frac{−\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{4}}\:\left(\sqrt{\mathrm{4}}\:+\mathrm{2}\right)}=−\frac{\mathrm{1}}{\mathrm{16}} \\ $$
Commented by Calculusboy last updated on 18/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$